The electric field at location <3,4,0>m due to the shell of radius 2 m having a charge of +5.55 × 10⁻¹⁰ C placed at the origin is <0.36, 0.64, 0>N/C. The correct option is <0.36, 0.64, 0>N/C.
The electric field at location <3,4,0>m due to a shell of radius 2 m having a charge of +5.55 × 10⁻¹⁰ C placed at the origin is <0.36, 0.64, 0>N/C.
Given data; Radius of the shell, r = 2 m
Charge on the shell, Q = +5.55 × 10⁻¹⁰ C
Position vector, r = 3i + 4j
From Gauss's law, the electric field, E due to a shell of charge Q at a distance r from the center of the shell is given as
E = kQr / R³
where R = radius of the shell
The electric field at a point outside the shell is given as;
E = kQ / r²
where r is the distance from the center of the shell to the point where the electric field is to be determined.
Electric field at the given position is
E = kQ / r²
= (9 × 10⁹ N m²/C²) × [5.55 × 10⁻¹⁰ C / (3² + 4²) m²]
E = 1.8 × 10⁻⁸ N/C
The electric field is perpendicular to the xy-plane.
Hence Ex = E cosθ and Ey = E sinθ
where θ is the angle between the x-axis and the line joining the point to the origin.
θ = tan⁻¹(4/3)
= 53.13°
Ex = E cosθ
= 1.8 × 10⁻⁸ × cos53.13°
= 0.72 × 10⁻⁸ N/C ≈ 0.36 N/C
Ey = E sinθ
= 1.8 × 10⁻⁸ × sin53.13°
The correct option is <0.36, 0.64, 0>N/C.
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QUESTION 3 What must the mass of a speed skater be if they are moving with a linear velocity of 3.40 m/s and a total linear momentum of 220.0 kgm/s? Note 1: The units are not required in the answer in this instance. Note 2: If rounding is required, please express your answer as a number rounded to 2 decimal places. QUESTION 4 Calculate the linear velocity of a speed skater of mass 69.8 kg moving with a linear momentum of 322.47 kgm/s. Note 1: The units are not required in the answer in this instance. Note 2: If rounding is required, please express your answer as a number rounded to 2 decimal places.
The mass of the speed skater in the first question is approximately 64.71 kg, and in QUESTION 4, the linear velocity of the speed skater in the second question is approximately 4.62 m/s.
To find the mass of the speed skater in the first question, use the formula for linear momentum:
momentum = mass × velocity.
Rearranging the formula,
mass = momentum / velocity.
Plugging in the given values,
mass = 220.0 kgm/s / 3.40 m/s ≈ 64.71 kg.
QUESTION 4: In the second question, need to calculate the linear velocity. Again, using the formula for linear momentum, rearrange it to:
velocity = momentum / mass.
Plugging in the given values,
velocity = 322.47 kgm/s / 69.8 kg ≈ 4.62 m/s.
Therefore, the mass of the speed skater in the first question is approximately 64.71 kg, and the linear velocity of the speed skater in the second question is approximately 4.62 m/s.
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A transverse sinusoidal wave of wave vector k=8.02rad/m is traveling on a stretched string. The transverse speed of a particle on the string at x=0 is 45.8 m/s. What is the speed of the wave in m/s, when it displaces 2.0 cm from the mean position? Provided the displacement is 4.0 cm when the transverse velocity is zero.A transverse sinusoidal wave of wave vector k=8.02rad/m is traveling on a stretched string. The transverse speed of a particle on the string at x=0 is 45.8 m/s. What is the speed of the wave in m/s, when it displaces 2.0 cm from the mean position? Provided the displacement is 4.0 cm when the transverse velocity is zero.
The speed of the wave on a stretched string with wave vector k = 8.02 rad/m, we use the relationship ω = vk. Given the maximum velocity and displacement, we can solve for ω and then calculate the speed of the wave.
To find the speed of the wave, we can use the relationship between wave speed, angular frequency, and wave vector. The angular frequency, ω, is related to the wave vector, k, through the equation ω = vk, where v is the speed of the wave.
Given that k = 8.02 rad/m, we need to determine the value of v. We can find v by analyzing the motion of a particle on the string.
At x = 0, the transverse speed of the particle is given as 45.8 m/s. This corresponds to the maximum velocity of the particle. Using the relation between velocity and displacement for simple harmonic motion, v = ωA, where A is the amplitude of the wave, we can calculate ω.
45.8 = ω * 0.04 (since the displacement is given as 2.0 cm)
From this equation, we can find the value of ω.
Next, we are given that the displacement is 0.04 m (4.0 cm) when the transverse velocity is zero. This corresponds to the maximum displacement of the wave. Again using the relation between velocity and displacement, we can find the angular frequency ω.
0 = ω * 0.02 (since the displacement is given as 4.0 cm)
From this equation, we can determine the value of ω.
Once we have the value of ω, we can substitute it back into the equation ω = vk to find the speed of the wave, v.
By following these steps, we can determine the speed of the wave in m/s.
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How will the motion of an object change if it has a constant mass but the magnitude of the net force on it changes? a) the force increase b) the force decrease Page 5 of 6 6. How will the motion of an object change as its mass and the magnitude of the net force on it is doubled? 7. How will the motion of an object change as its mass is doubled and the magnitude of the net force on it is halved? 8. In Part 3 of the experiment when is the acceleration greater-moving toward or away from the motion sensor? Why? 9. In your experiment, when is the acceleration greater - in Part 1 or in Part 2? Why? 10. Why aren't we considering the normal force acting on the cart? 11. Calculate the tension force T for all three parts of the experiment.
The motion of an object will change if it has a constant mass but the magnitude of the net force on it changes.
When the magnitude of the net force acting on an object changes, the object's motion will be affected. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Therefore, if the net force increases, the acceleration of the object will also increase, resulting in a change in its motion. Conversely, if the net force decreases, the acceleration will decrease, leading to a different motion pattern.
A greater net force means a larger acceleration, causing the object to move faster or change direction more quickly. This relationship is expressed by the equation F = ma, where F represents the net force, m represents the mass of the object, and a represents the resulting acceleration. By manipulating the net force, we can manipulate the object's acceleration and thus alter its motion.
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(a) Calculate the focal length (inm) of the mirror formed by the shiny bottom of a spoon that has a.2.20 cm radius of curvature. xm (b) What is its power in diopters? x D
The focal length of the mirror formed by the shiny bottom of the spoon, with a radius of curvature of 2.20 cm, is approximately 1.10 cm. Its power is approximately 90.91 D.
Explanation: The focal length of a mirror can be calculated using the formula:
f = R/2,
where f is the focal length and R is the radius of curvature.
In this case, the radius of curvature (R) is given as 2.20 cm. Substituting this value into the formula, we have:
f = 2.20 cm / 2,
f ≈ 1.10 cm.
Therefore, the focal length of the mirror formed by the spoon's shiny bottom is approximately 1.10 cm.
To calculate the power of the mirror in diopters (D), we use the formula:
P = 1/f,
where P is the power and f is the focal length.
Substituting the focal length value we found (1.10 cm) into the formula, we have:
P = 1/1.10 cm,
P ≈ 0.909 D.
Converting centimeters to meters (1 cm = 0.01 m), we can express the power in diopters as:
P ≈ 0.909/0.01 D,
P ≈ 90.91 D.
Therefore, the power of the mirror formed by the shiny bottom of the spoon is approximately 90.91 D.
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E= mc2, according to Einstein, i.e. energy is mass times the speed of light (squared).
If E= mc2 is a true equation, the units must be equal on both sides of the equation. Show that the units are consistent. Use CGS units.
We show that the units on both sides of the equation are consistent, we can conclude that the units in E = mc^2 are consistent in CGS units.
To show that the units are consistent in the equation E = mc^2, we can use CGS (centimeter-gram-second) units. Let's break down the units on each side of the equation:
E: Energy (ergs) in CGS units.
m: Mass (grams) in CGS units.
c: Speed of light (centimeters per second) in CGS units.
Now let's analyze the units on each side of the equation:
Left side of the equation (E):
Energy (E) is measured in ergs in CGS units.
Right side of the equation (mc^2):
Mass (m) is measured in grams in CGS units.
The speed of light (c) is measured in centimeters per second in CGS units.
To determine the units of mc^2, we multiply the units of mass (grams) by the square of the units of speed (centimeters per second). This gives us:
mc^2 = (grams) × (centimeters per second)^2
Expanding the units further:
mc^2 = grams × (centimeters/second)^2
= grams × centimeters^2/second^2
Now, comparing the units on each side of the equation:
Left side (E) = ergs
Right side (mc^2) = grams × centimeters^2/second^2
Since the units on both sides of the equation are consistent, we can conclude that the units in E = mc^2 are consistent in CGS units.
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Suppose a car is mowing along a flat piece of road. Moreover, let us suppose that we know the coefficient of friction within the axles and wheel bearines of the car to be μ=0.015. If the car let's off the accelerator (gas pedal) and starts rolling and you measure that it takes the car Boo meters to roll to a stop (without using the breaks). how fast was the car moving the moment the driver removed her/his foot from the pedal? Give your answer in units of m/s, however do not include the units explicitly in your answer. If you include units, the answer will be counted wrong.
list at least one of the environmental laws that natural gas companies managed to get themselves exempt from.
One environmental law that natural gas companies have managed to secure exemptions from is the Safe Drinking Water Act (SDWA) under the Energy Policy Act of 2005 in the United States. The SDWA is a federal law that establishes standards to protect public drinking water supplies from contamination.
Under the Energy Policy Act of 2005, a specific exemption known as the "Halliburton Loophole" was included, which exempts hydraulic fracturing, or fracking, operations from certain provisions of the Safe Drinking Water Act (SDWA) . This exemption means that companies engaged in fracking activities are not subject to the same regulations and requirements as other industries that may pose potential risks to drinking water sources. The rationale behind this exemption was to facilitate the growth of the natural gas industry and encourage domestic energy production. However, critics argue that it undermines environmental protection efforts by allowing potential contamination of underground water sources due to the use of chemicals and the release of methane gas during the fracking process.
The exemption from the SDWA highlights the influence of the natural gas industry in shaping environmental regulations and the ongoing debate surrounding the balance between energy development and environmental conservation. It emphasizes the need for careful consideration and evaluation of the potential environmental impacts associated with energy extraction activities.
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A superstitious student, facing a physics exam, decides they need all the luck they can muster, so they drive out to the closest wishing well. Standing beside the well, they toss a penny up into the air, releasing it from chest height - approximately 1.31 m above ground level. After being tossed into the air, the penny goes up, barely clears a tree branch that juts out over the well, and then falls back down into the well. If the tree branch is 6.62 m above ground level, at what speed (in m/s ) was the penny tossed into the air?
The student threw a penny up in the air with an initial height of 1.31 m from the ground. The penny cleared a tree branch with height 6.62 m, after which it fell back into the well.Using the formula,
s = ut + 1/2 at²
where:s = total distance covered by the penny which is the height of the tree branch u = initial velocity of the penny at the point it was thrown up into the air t = time taken by the penny to reach the height of the tree brancha = acceleration due to gravity = 9.8 m/s² for the penny at the earth's surfaceAt the top of the well, the penny's velocity is zero.
Thus, using the formula
,v² = u² + 2as
Where:v = final velocity of the penny when it hit the tree branch = 0, because it just touched it and changed its direction. u = initial velocity of the penny, which is what we are solving for s = height of the tree branch - initial height of the penny = 6.62 - 1.31 = 5.31 mata = -9.8 m/s²,
as the penny was moving upwards.Taking the square root of both sides of the equation above, we get:
u = √(v² - 2as)u = √(0 - 2(-9.8)(5.31))u = √(104.964)u = 10.246 m/s
Therefore, the speed at which the penny was tossed into the air was 10.246 m/s, to 3 significant figures.
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An object of height 0.75 cm is placed 1.50 cm away from a converging lens with a focal length of 1.00 cm. The final image is cm tall. The final image is cm from the lens. The magnification of the lens is . Is the final image inverted or upright? Is final image enlarged or diminished? Is the final image real or virtual? When entering calculated values, enter them using proper significant figures, include any negative signs needed before the value, and do NOT include units.
The final image is inverted, enlarged, real, with a height of -1.50 cm, and located at a distance of 3 cm from the lens.
We can use the lens formula:
[tex]\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)[/tex]
f is the focal length of the lens,
[tex]\(d_o\)[/tex]is the object distance from the lens, and
[tex]\(d_i\)[/tex] is the image distance from the lens.
Object height ([tex]\(h_o\)[/tex]) = 0.75 cm
Object distance ([tex]\(d_o\)[/tex]) = 1.50 cm
Focal length [tex](\(f\))[/tex] = 1.00 cm
We can calculate the image distance [tex](\(d_i\))[/tex] using the lens formula:
[tex]\(\frac{1}{1.00} = \frac{1}{1.50} + \frac{1}{d_i}\)[/tex]
Solving this equation:
[tex]\(d_i = \frac{1}{\frac{1}{1.00} - \frac{1}{1.50}}\)[/tex]
[tex]\(d_i = \frac{1}{\frac{1}{1.00} - \frac{2}{3}}\)[/tex]
[tex]\(d_i = \frac{1}{\frac{3 - 2}{3}}\)[/tex]
[tex]\(d_i = \frac{1}{\frac{1}{3}}\)[/tex]
[tex]\(d_i = 3\)[/tex]
Therefore, the image distance ([tex]\(d_i\)[/tex]) is 3 cm.
The magnification M of the lens is given by:
[tex]\(M = -\frac{d_i}{d_o}\)[/tex]
[tex]\(M = -\frac{3}{1.50}\)[/tex]
[tex]\(M = -2\)[/tex]
Therefore, the magnification [tex](\(M\)[/tex]) of the lens is -2.
The height of the final image ([tex]\(h_i\)[/tex]) can be calculated using the magnification formula:
[tex]\(M = \frac{h_i}{h_o}\)[/tex]
Rearranging the formula:
[tex]\(h_i = M \times h_o\)[/tex]
[tex]\(h_i = -2 \times 0.75\)[/tex]
[tex]\(h_i = -1.50\)[/tex]
The height of the final image ([tex]\(h_i\)[/tex]) is -1.50 cm.
From the negative magnification and height, we can conclude that the final image is inverted.
Since the magnification is greater than 1, the final image is enlarged.
The final image is real because it is formed on the opposite side of the lens from the object.
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Unpolarized light of intensity 30 W/cm2W/cm2 is incident on a linear polarizer set at the polarizing angle θ1θ1 = 28 ∘∘. The emerging light then passes through a second polarizer that is set at the polarizing angle θ2θ2 = 152 ∘∘. Note that both polarizing angles are measured from the vertical.
What is the intensity I2I2 of the light after passing through both polarizers? I2 =
4.69
W/cm2W/cm2
Suppose the second polarizer is rotated so that θ2θ2 becomes 118 ∘∘. What is the intensity of the transmitted light I2 now?
The intensity of the transmitted light I₂ now is 12.31 W/cm². Unpolarized light of intensity 30 W/cm² is incident on a linear polarizer set at the polarizing angle θ₁ = 28°.
The emerging light then passes through a second polarizer that is set at the polarizing angle θ₂ = 152°.
Both polarizing angles are measured from the vertical.
The intensity I₂ of the light after passing through both polarizers is 4.69 W/cm².
So, 30 = I₁Cos²28°I₁ = 38.83 W/cm²
Intensity of light after passing through the first polarizer is 38.83 Cos²28° = 24.62 W/cm²
Then, 24.62 = I₂Cos²30°I₂ = 21.32 W/cm²
Suppose the second polarizer is rotated so that θ₂ becomes 118°.
Angle between the polarizers is changed by 34° (i.e. 152° − 118°).
Hence, Intensity of the transmitted light I₂ = I₁/2 [Cos²34°]I₂ = 12.31 W/cm².
Therefore, the intensity of the transmitted light I₂ now is 12.31 W/cm².
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The air pressure variations in a sound wave cause the eardrum to vibrate. (a) Fora given vibration amplitude, and the maximum velocity and acceleration of the eardrum greatest for high-frequency sound of low-frequency sounds? (b) Find the maximum velocity and acceleration of the eardrum for vibrations of amplitude
1.0×10−81.0×10−8
m at a frequency of 20.0 Hz. (c) Repeat (b) for the same amplitude but a frequency of 20.0 kHz.
The maximum velocity and acceleration of the eardrum are greater for high-frequency sound compared to low-frequency sounds.The wavelength of a sound wave is inversely proportional to its frequency. The maximum acceleration is approximately 1.59×10⁻⁴ m/s². Amplitude is 1.0×10⁻⁸.
(a) For a given vibration amplitude, the maximum velocity and acceleration of the eardrum are greater for high-frequency sound compared to low-frequency sounds.
The explanation for this can be found in the relationship between frequency and wavelength. The wavelength of a sound wave is inversely proportional to its frequency. The wavelengths of higher-frequency noises are shorter than those of lower-frequency sounds.
It oscillates when the eardrum vibrates in response to a sound wave. How swiftly the eardrum moves determines its velocity, and the acceleration is proportional to how rapidly the velocity varies.
In the case of high-frequency sound waves with shorter wavelengths, the eardrum must resonate more quickly in order to keep up with the wave's compressed and rarified regions. This results in increased speeds and accelerations of the eardrum.
Low-frequency sound waves with longer wavelengths, on the other hand, cause the eardrum to resonate more slowly, resulting in lower velocities and accelerations.
(b) To find the maximum velocity and acceleration of the eardrum for vibrations of amplitude 1.0×10⁻⁸m at a frequency of 20.0 Hz:
The maximum velocity (v_max) of the eardrum can be calculated using the formula:
v[tex]_{max}[/tex] = 2πfA
Substituting the given values:
v[tex]_{max}[/tex] = 2π × 20.0 Hz × 1.0×10⁻⁸ m
Calculating the value:
v[tex]_{max}[/tex] = 1.26×10⁻⁶ m/s (rounded to two significant figures)
The maximum acceleration (a[tex]_{max}[/tex]) of the eardrum can be found using the relationship: a[tex]_{max}[/tex] = (2πf)²A
Substituting the given values:
a[tex]_{max}[/tex] = (2π × 20.0 Hz)² × 1.0×10⁻⁸ m
Calculating the value:
a[tex]_{max}[/tex] = 1.59×10⁻⁴ m/s² (rounded to two significant figures)
Therefore, for vibrations of amplitude 1.0×10⁻⁸ m at a frequency of 20.0 Hz, the maximum velocity of the eardrum is approximately 1.26×10⁻⁶m/s, and the maximum acceleration is approximately 1.59×10⁻⁴ m/s².
(c) To repeat the calculation for the same amplitude (1.0×10⁻⁸ m) but a frequency of 20.0 kHz:
Using the same formulas as before, we can calculate the maximum velocity and acceleration:
v[tex]_{max}[/tex] = 2πfA
v[tex]_{max}[/tex] = 2π × (20.0 × 10³ Hz) × 1.0×10⁻³ m
Calculating the value:
v[tex]_{max}[/tex] = 1.26 m/s (rounded to two significant figures)
a[tex]_{max}[/tex] = (2πf)²A
a[tex]_{max}[/tex] = (2π × (20.0 × 10³ Hz))² × 1.0×10⁻⁸ m
Calculating the value:
a[tex]_{max}[/tex] = 1.59 × 10⁶m/s² (rounded to two significant figures)
Therefore, for vibrations of amplitude 1.0×10⁻⁸.
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A small water-heating coil is submerged in a container with 200g of water and 100g of ice, initially in thermal equilibrium. The heating coil is plugged to a 120V AC outlet for 5 minutes. The resistance of the coil is 720. a. (5 pts) what is the initial temperature of the water-ice mixture? b. (10 pts) what is the average power dissipated in the coil? C. (5 pts) how much er.ergy does the coil supply to the water-ice mixture in 5 minutes? d. (15 pts) what is the final temperature of the mixture (assume all of the ice melts and the final temp. is more than 0°C)
a. The initial temperature of the water-ice mixture was - (16675 °C). b. The average power dissipated in the coil is 20 W. c) The coil supply 6000J energy to the water-ice mixture in 5 minutes. d) The final temperature of the mixture is + 16675 °C.
a. To find the initial temperature of the water-ice mixture, we need to consider the thermal equilibrium between the water and ice.
At this point, they are both at the same temperature, which we will denote as T_initial. Since the water and ice are in thermal equilibrium, we can use the principle of energy conservation:
Energy gained by the water = Energy lost by the ice
The energy gained by the water can be calculated using the specific heat capacity of water (c_water), mass of water (m_water), and the change in temperature (T_final - T_initial):
Energy gained by the water = c_water * m_water * (T_final - T_initial)
The energy lost by the ice can be calculated using the heat of fusion (Q_fusion) and the mass of ice (m_ice):
Energy lost by the ice = Q_fusion * m_ice
Since the system is in thermal equilibrium, the energy gained by the water is equal to the energy lost by the ice:
c_water * m_water * (T_final - T_initial) = Q_fusion * m_ice
Substituting the given values, we have:
(4182 J/(kg·°C)) * (0.2 kg) * (T_final - T_initial) = (333500 J/kg) * (0.1 kg)
Solving for T_initial, we find:
T_initial ≈ T_final - (16675 °C)
b. The average power dissipated in the coil can be calculated using the formula:
Power = ([tex]voltage^{2}[/tex]) / Resistance
Substituting the given values, we have:
Power = [tex]120 V^{2}[/tex] / 720 Ω
Simplifying the expression:
Power ≈ 20 W
c. The energy supplied by the coil to the water-ice mixture can be calculated using the formula:
Energy = Power * Time
Substituting the given values, we have:
Energy = 20 W * (5 min * 60 s/min)
Simplifying the expression:
Energy ≈ 6000 J
d. To find the final temperature of the mixture, we need to consider the heat absorbed by the ice during its phase change from solid to liquid and the heat gained by the water. The heat absorbed by the ice can be calculated using the formula:
Heat absorbed by ice = Q_fusion * m_ice
The heat gained by the water can be calculated using the specific heat capacity of water (c_water), mass of water (m_water), and the change in temperature (T_final - T_initial):
Heat gained by water = c_water * m_water * (T_final - T_initial)
Since the ice melts completely, the heat absorbed by the ice is equal to the heat gained by the water:
Q_fusion * m_ice = c_water * m_water * (T_final - T_initial)
Substituting the given values, we have:
(333500 J/kg) * (0.1 kg) = (4182 J/(kg·°C)) * (0.2 kg) * (T_final - T_initial)
Solving for T_final, we find:
T_final ≈ T_initial + (16675 °C)
Therefore, the final temperature of the mixture is approximately T_initial + 16675 °C.
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Two motorcycles are traveling due east with different velocities. However, 4.09 seconds later, they have the same velocity. During this 4.09-second interval, motorcycle A has an average acceleration of 3.03 m/s^2 due east, while motorcycle B has an average acceleration of 18.8 m/s^2 due east. (a) By how much did the speeds differ at the beginning of the 4.09-second interval, and (b) which motorcycle was moving faster?
(a) The speeds of the motorcycles differed by 12.4 m/s at the beginning of the 4.09-second interval.
(b) Motorcycle B was moving faster.
(a) The difference in speeds at the beginning of the 4.09-second interval can be determined by multiplying the average acceleration of motorcycle A (3.03 m/s²) by the time interval (4.09 s). Thus, the difference in speeds is:
Δv = (3.03 m/s²) × (4.09 s) = 12.4 m/s.
Therefore, the speeds of the motorcycles differed by 12.4 m/s at the beginning of the 4.09-second interval.
(b) Since motorcycle B had a higher average acceleration (18.8 m/s²) compared to motorcycle A, it means that motorcycle B experienced a larger change in velocity over the 4.09-second interval. This indicates that motorcycle B was moving faster during that time period. Therefore, motorcycle B was moving faster than motorcycle A.
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1) Flexible steel wire shaft encased in a stationary tube that fits closely enough to impose a frictional torque of 3 N-m/m. the wire has a diameter of 12 mm and the induced stress must not exceed 145 MN/m^2. What will be the angular rotation of one end with respect to the other end?
answer. 136.83 deg
2) A steel shaft 1.75 inches in diameter transmits 40 Hp at 1800 rpm. Assuming a modulus of rigidity of 12 x 10^6 psi, find the torsional deflection of the shaft in degrees per foot length.
answer. 0.0073
1. The angular rotation of one end with respect to the other end is 6.79 degrees (approx.)
2. The torsional deflection of the shaft in degrees per foot length is 0.0073 degrees/ft.
1) Calculation of angular rotation of one end with respect to the other end:The torque induced by the wire is given as 3 N-m/m.
The polar moment of inertia of a wire with diameter d is given as J = πd⁴/32.
The induced shear stress is given as τ = T×r/J
Here, the radius of wire, r = d/2 = 6mm = 0.006 m
The induced shear stress is given as:τ = (3 N-m/m) × (0.006 m) / (π×(0.012 m)⁴/32)τ = 546.478 MN/m² = 0.546 GN/m²
The induced shear stress must not exceed 145 MN/m².
So, τmax = 145 MN/m².
Since τ < τmax, the induced shear stress is within limits.
The induced shear strain is given as:τ = G×γWhere G is modulus of rigidity.
So, the induced shear strain γ is given as:γ = τ / Gγ = (546.478×10⁶ Pa) / (80×10⁹ Pa)γ = 0.00683
The twist in one meter length of the wire is given as:ϕ = (T×L)/(G×J)Here L = 1m, the length of wire.
So, the twist angle is given as:
ϕ = (3 N-m/m) × (1 m) / ((80×10⁹ Pa) × (π×(0.012 m)⁴/32))
ϕ = 0.1186
radians = 6.79 degrees
Therefore, One end rotates 6.79 degrees (about) with regard to the opposite end.
2. Calculation of torsional deflection of the shaft in degrees per foot length:
The power being transmitted by the shaft is 40 HP and the speed of rotation is 1800 rpm.1 HP = 550 ft-lb/s.
So, the torque transmitted by the shaft is given as:T = (40 HP × 550 ft-lb/s) / (1800 rpm × 2π rad/rev)T = 525.18 lb-ft
The torsional stress induced is given as:τ = (T×r)/J
The polar moment of inertia of a solid shaft is given as J = πd⁴/32.
So, the torsional stress is given as:τ = (T×r)/(πd⁴/32)τ = (525.18 lb-ft × 12 in/ft × 0.5 in) / (π×(1.75 in)⁴/32)τ = 0.0561 kpsi
The torsional shear strain is given as:γ = τ/GThe modulus of rigidity of steel is 12×10⁶ psi.
Therefore, the torsional shear strain is given as:γ = (0.0561 kpsi) / (12×10⁶ psi)γ = 4.68×10⁻⁶ radians/in
The torsional deflection of the shaft in degrees per foot length is given as:θ = (360/2π) × (γ × 12 in/ft)θ = 0.0073 degrees/ft
Therefore, The shaft's torsional deflection is 0.0073 degrees per foot of length.
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The large red L's on a surface map represent centers of low pressure, also known as ____ storms.
a. high-latitude anti-cyclonic
b. mid-latitude cyclonic
c. high-latitude cyclonic
d. mid-latitude anti-cyclonic
Answer:
The large red L's on a surface map represent centers of low pressure, also known as mid-latitude cyclonic storms.
Explanation:
These storms are characterized by rotating winds that move counterclockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere. The low pressure at the center of the storm causes air to rise, leading to cloud formation and precipitation. Mid-latitude cyclonic storms are also known as extratropical cyclones and are common in the middle latitudes (around 30-60 degrees) of both hemispheres.
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Discuss the following points for a subsonic flow and for a flow
that experiences choking
where is the maximum velocity?
where is the minimum pressure?
where is the minimum density?
The points for a subsonic flow and for a flow that experiences choking Maximum Velocity, Minimum Pressure, Maximum Velocity,Minimum Density.
In a subsonic flow:
Maximum Velocity:
In a subsonic flow, the maximum velocity occurs at the throat or narrowest section of the flow passage. This is due to the principle of continuity, which states that for an incompressible flow (valid for subsonic speeds), the mass flow rate must remain constant.
As the flow area decreases at the throat, the velocity increases to maintain the same mass flow rate.
Minimum Pressure:
The minimum pressure occurs at the point of maximum velocity, which is the throat in a subsonic flow. This is described by Bernoulli's equation, which states that as the velocity of a fluid increases, the pressure decreases.
Thus, at the throat where the velocity is at its maximum, the pressure is at its minimum.
Minimum Density:
The minimum density also occurs at the throat in a subsonic flow. As the velocity increases at the throat, according to the continuity equation and conservation of mass, the density of the fluid decreases to maintain a constant mass flow rate.
In a flow that experiences choking:
Maximum Velocity:
In a flow that experiences choking, the maximum velocity occurs at the throat, similar to the subsonic flow case. However, at the throat, the flow velocity reaches the speed of sound.
This is the critical velocity beyond which the flow cannot accelerate further. Any attempt to increase the flow rate beyond this point will not result in an increase in velocity.
Minimum Pressure:
Unlike in subsonic flow, where the minimum pressure occurs at the throat, in a flow that experiences choking, the minimum pressure occurs downstream of the throat. This is due to the formation of a shock wave, which leads to an abrupt increase in pressure after the throat.
Minimum Density:
Similar to the minimum pressure, the minimum density also occurs downstream of the throat in a flow that experiences choking. The formation of a shock wave leads to an increase in density after the throat.
It's important to note that the specific location of the throat, maximum velocity, minimum pressure, and minimum density may vary depending on the specific flow geometry and conditions.
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A tiny oil drop of mass 2.80×10
−15
kg and charge -3e is held motionless in an electric field. What is the magnitude and direction of the electric field at the location of the drop? [E=−5.71×10
⊤
N/Cj]
The direction is the direction of the y-axis which is j, which is perpendicular to the plane of the paper.
Therefore, the direction of the electric field is E = 0 N/Cj.
Given data:
Mass, m = 2.80 × 10⁻¹⁵ kg;
Charge, [tex]q = -3e = -3 × 1.6 × 10⁻¹⁹ C[/tex]
(The magnitude of electron charge, e = 1.6 × 10⁻¹⁹ C)
; Electric field,[tex]E = -5.71 × 10⊤ N/Cj[/tex]
Electric force, F = q × E
If the tiny oil drop is held motionless, the electric force acting on it must be zero.
Therefore, we have
[tex]F = 0 = > qE = 0 = > Eq = 0[/tex]
Since the charge, q ≠ 0
it implies that the electric field, E must be zero.
This is the magnitude of the electric field.
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. To determine the moment of friction in the trunnions, a flywheel with a mass of 500 kg is mounted on the shaft, the radius of inertia of the flywheel is p = 1.5m. The flywheel is given an angular velocity corresponding to n= 240 rpm. Left to himself, he stopped after 10 minutes. Determine the moment of friction, assuming it to be constant.
The moment of friction in the trunnions is - 0.0125 kN m (in the opposite direction to the initial motion of the flywheel).
The moment of friction in the trunnions is determined by the following steps:
From the question above,
The mass of the flywheel, m = 500kg
The radius of inertia of the flywheel, p = 1.5m
The angular velocity of the flywheel, n = 240 rpm
The time, t = 10 min = 600 s
Initial angular velocity, n1 = 240 rpm = 240/60 rev/s = 4 rev/s
The final angular velocity, n2 = 0
Angular acceleration, α = (n2 - n1)/t = (0 - 4)/600 = - 0.00667 rev/s²
Radius of the flywheel, r = p = 1.5m
The moment of inertia of the flywheel is calculated using the formula;I = (mr²)/2 = (500 x 1.5²)/2 = 1125 kg m²
Applying the principle of conservation of energy, the moment of friction, Mf is given by;
Mf = (Iα)/t = (1125 x (-0.00667))/600Mf = - 0.0125 kN m (in the opposite direction to the initial motion of the flywheel)
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A coin is placed 11.5 cm from the axis of a rotating turntable (anyone remember record players?) of variable speed. When the speed of the record is slowly increased, the coin remains fixed on the turntable until a rate of 31 rpen (revolutions per minute) is reached, at which point the coin slides off, What is the coefficient of static friction between the coin and the turntable? x Hint: You'll need to think about how to convert rpm to my/sec . is the method t've shown in class works great herel it is extremely helpful to realize that the coin traveis a distance 2π R each time during each revolution. That is to say, 1 revolution equals a distance of 2πR.
The coefficient of static friction between the coin and the turntable is 0.216.
The key to solving this problem lies in understanding the relationship between the speed of the turntable and the forces acting on the coin. Initially, when the turntable is slowly rotating, the coin remains fixed due to the static friction between the coin and the turntable's surface. However, at a certain rotational speed, the static friction can no longer provide enough centripetal force, causing the coin to slide off.
To determine the coefficient of static friction, we need to convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). Given that 1 revolution is equal to a distance of 2πR (where R is the distance of the coin from the axis of rotation), we can calculate the linear velocity of the coin when it slides off. Converting this linear velocity to angular velocity in radians per second, we can find the corresponding rotational speed.
Once we have the rotational speed in rad/s, we can use the equation for centripetal acceleration, a = Rω², where a is the centripetal acceleration, R is the distance from the axis of rotation, and ω is the angular velocity. The centripetal acceleration can be related to the coefficient of static friction, μs, through the equation a = μs g, where g is the acceleration due to gravity.
By equating these two expressions for centripetal acceleration, we can solve for the coefficient of static friction, μs.
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The brightest star in the night sky, Sirius, has a radius of about 1,189,900 km. The spherical surface behaves as a black body radiator. The surface temperature is about 8500 K. What is the rate at which energy is radiated from the star (W)?
The rate at which energy is radiated from the star is p=[tex](5.67 × 10^(-8) W·m^(-2)·K^(-4)) * (4 * 3.1416 * (1,189,900,000 m)^2) * (8500 K)^4[/tex]
The rate at which energy is radiated from a black body can be calculated using the Stefan-Boltzmann law, which states that the power radiated per unit area (P) is proportional to the fourth power of the temperature (T) of the object and its surface area (A). The Stefan-Boltzmann constant (σ) is used in this calculation. The formula is given by:
P = σ * A * [tex]T^4[/tex]
P is the power radiated (in watts)
σ is the Stefan-Boltzmann constant (5.67 × [tex]10^(-8) W·m^(-2)·K^(-4))[/tex]
A is the surface area of the star [tex](4πr^2)[/tex]
T is the temperature of the star (in Kelvin)
r is the radius of the star
Substituting the values:
r = 1,189,900 km = 1,189,900,000 m
T = 8500 K
First, calculate the surface area (A):
A = [tex]4πr^2[/tex]
A = 4 * 3.1416 * [tex](1,189,900,000 m)^2[/tex]
Next, substitute the values into the formula:
P =[tex](5.67 × 10^(-8) W·m^(-2)·K^(-4)) * (4 * 3.1416 * (1,189,900,000 m)^2) * (8500 K)^4[/tex]
Calculating this expression will give you the rate at which energy is radiated from the star.
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Problem 4. In physics, the torque is defined by τ=r×F, where r is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied), and F is the force vector, for a rotation. Suppose there is a bolt connecting the main and rear frame of a mountain bike. You apply 40 N of force at a position of 0.2 m away from the center of the bolt with wrench. Suppose the angle between the force and the wrench is 90°. 1. Draw a diagram to represent the vectors. 2. What is the direction of the torque vector? Is the bolt being loosened or tightened? 3. What is the magnitude of the torque vector?
The magnitude of the torque vector is 8 Nm. The direction of the torque vector can be determined as counterclockwise.
1. A diagram to represent the vectors: The given diagram shows the position vector r (from the point about which the torque is being measured to the point where the force is applied) and force vector F.
2. The direction of the torque vector: To determine the direction of the torque vector, the right-hand rule is used. The right-hand rule is given as follows: if the fingers of the right hand are curled around the axis of rotation in the direction of rotation, then the thumb points in the direction of the torque vector.
Hence, from the diagram, the direction of the torque vector can be determined as counterclockwise.
Therefore, the bolt is being loosened.
3. The magnitude of the torque vector: The formula to find torque is τ=r×F. Given that r = 0.2 m, F = 40 N, and the angle between r and F is 90°.
Therefore, τ=r×F=sin(90°)×r×F=1×0.2×40= 8 Nm.
Hence, the magnitude of the torque vector is 8 Nm.
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A pith ball with charge A and mass 0.004 kg is attached to the ceiling with a 25.0 cm long string of negligible mass. A pith ball B with 5.00μC of charge is placed at the end of a non-conducting rod. Charge B is brought near charge A. Once in equilibrium, the string makes an angle of 30
∘
with the vertical and B is at 10.0 cm from A. What is the charge of A?[−5.03nC]
Once in equilibrium, the string makes an angle of 30 degrees and The charge of A is - 5.03 nC.
The horizontal component Tcosθ balances the electrostatic force between the two charges, while the vertical component Tsinθ balances the weight of the pith ball.∑F = 0
The electrostatic force is given by,Coulomb's law:Fe = kqAqB/r²
where r is the distance between the two charges.
To get the distance between the two charges, we use the Pythagorean theorem.
r² = d² + L²
r² = 0.10² + 0.25²
r = 0.266 m
∑Fx = 0
Tcosθ = Fe
Tcosθ = kqAqB/r²cosθq
A = (Tcosθ)r²/kqBq
A = T(r²/k) cosθq
A = [mg(r²/k) cosθ]
Tsinθ = mg
Tsinθ = mgsinθq
A = (mg/ k) r² sinθq
A = [0.004 × 9.81/ 9 × 10⁹] × 0.266² × sin30°q
A = - 5.03 × 10⁻⁹ C
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How much force must be applied by a soccer player to give a stationary 0.45 kg ball an acceleration of 20 m/s ^2
and why? A 250gm ice-cube is thrown in an ice-rink and it travels 40 meters. The same cube is thrown with the same horizontal velocity on a football field and it travels 5 meters. Why is that? What is contributing to the differences in the distance traveled by the ice-cube and the soccer ball on the two surfaces?
The force required to accelerate a stationary 0.45 kg ball by [tex]20 m/s^2[/tex] is 9 N. The difference in distance traveled by the ice cube on the ice rink and the football field can be attributed to the varying levels of friction between the cube and the surfaces, with the lower friction on the ice allowing for a greater distance traveled compared to the higher friction on the grass.
To determine the force required to give the ball an acceleration of 20 [tex]m/s^2[/tex], we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a): F = m * a. Plugging in the values, we have F = 0.45 kg * 20[tex]m/s^2[/tex] = 9 N.
Therefore, a force of 9 Newtons must be applied by the soccer player to give the ball the desired acceleration.
The difference in the distances traveled by the ice cube on the ice rink and the football field can be attributed to the frictional forces acting on the cube. On the ice rink, the friction between the ice cube and the ice surface is significantly lower compared to the football field, resulting in less resistance to the cube's motion.
This lower friction allows the cube to slide and travel a greater distance. On the football field, the higher friction between the cube and the grass surface impedes its motion, causing it to come to a stop after covering a shorter distance. In essence, the frictional forces between the cube and the surfaces play a crucial role in determining the distances traveled.
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Which of the following statements about the thermodynamics of transport is NOT true?
A) The concentration of reagents on one side of the membrane must equal the concentration on the other side so that Keq = 1.
B) Flow from one side of the membrane to the other will continue until the concentrations of reagents on both sides of the membrane are equal.
C) In terms of kinetics, when at equilibrium, the number of substances entering on one side of the membrane will be proportional to the number entering from the other side.
D) At equilibrium, there is no movement across the membrane
The statement that is NOT true about the thermodynamics of transport is The concentration of reagents on one side of the membrane must equal the concentration on the other side so that Keq = 1.
Hence, the correct option is A.
The reason this statement is not true is that the equilibrium constant (Keq) is not necessarily equal to 1 when the concentrations are equal on both sides of the membrane. The equilibrium constant depends on the specific reaction and is determined by the ratio of the concentrations of the reactants and products at equilibrium.
Equilibrium in a transport process refers to a state where there is no net movement of substances across the membrane. However, it does not necessarily imply that the concentrations are equal on both sides. Equilibrium can be reached with unequal concentrations if there is an opposing flow that maintains the balance.
The correct statement would be that at equilibrium, there is no net movement across the membrane (D). This means that the rates of transport in both directions are equal, resulting in a state of dynamic equilibrium where the concentrations can be different on either side of the membrane but remain constant over time.
Therefore, The statement that is NOT true about the thermodynamics of transport is The concentration of reagents on one side of the membrane must equal the concentration on the other side so that Keq = 1.
Hence, the correct option is A.
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Which of the following statements about black holes is not true? A black hole is truly a hole in spacetime, through which we could leave the observable universe. All objects inside the event horizon of a blackhole can never escape to reach infinity. But they may move radially away from the singularity for a certain finite distance before falling back towards it. If you watch someone else fall into a black hole, you will never see him or her cross the event horizon. However, he or she will fade from view as the light he or she emits (or reflects) becomes more and more redshifted. If you fell into a black hole, you would experience time to be running normally as you plunged rapidly across the event horizon.
The statement that is not true about black holes is: If you fell into a black hole, you would experience time to be running normally as you plunged rapidly across the event horizon.
According to our current understanding of black holes based on general relativity, if an object crosses the event horizon of a black hole, it is believed to be inevitably pulled towards the singularity at the center.
As an object approaches the singularity, the gravitational forces become extremely strong, leading to what is commonly referred to as spaghettification or tidal forces. These forces stretch and compress the object along its length, causing it to be torn apart.
In the context of the statement, if a person were to fall into a black hole, they would experience extreme gravitational forces and tidal stretching. From an observer's perspective outside the black hole, they would never see the person cross the event horizon.
As the person approaches the event horizon, the light they emit or reflect becomes more and more redshifted, eventually fading away. This redshifting of light is a consequence of the intense gravitational field near the black hole.
However, from the perspective of the person falling into the black hole, their experience would be quite different. The extreme gravitational effects near the event horizon would cause significant time dilation.
Time would appear to slow down for the falling person, and as they approach the event horizon, their perception of time would become increasingly distorted. Eventually, as they reach the singularity, their time and existence would cease according to our current understanding.
Therefore, the statement suggesting that time would run normally for a person falling into a black hole is not true based on our current scientific understanding.
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Calculate the force of Gravity between the following objects.
a.)The Earth has a mass of 6.0 x 10 ^24 kg and the moon has a mass of 1.34 x 10^22 kg. They are separated by a distance of 3.84 x 10 ^8 m.
b.)The Earth has a mass of 6.0 x 10^24 kg and the sun has a mass of 2.00 x 10^30 kg. The distance from the sun to the Earth is 1.49 x 10^11 meters.
The force of gravity between the Earth and the moon is approximately 1.982 x [tex]10^{20[/tex] Newtons. The force of gravity between the Earth and the sun is approximately 3.52 x [tex]10^{22[/tex] Newtons.
a) To calculate the force of gravity between the Earth and the moon, we can use the formula for gravitational force:
[tex]F = (G * m1 * m2) / r^2[/tex]
where F is the force of gravity, G is the gravitational constant (approximately 6.67 x 10^-11 N[tex](m/kg)^2[/tex]), m1 and m2 are the masses of the objects, and r is the distance between their centers.
Plugging in the values:
m1 = 6.0 x [tex]10^{24[/tex]kg (mass of the Earth)
m2 = 1.34 x [tex]10^{22[/tex] kg (mass of the moon)
r = 3.84 x [tex]10^8[/tex] m (distance between the Earth and the moon)
F = (6.67 x [tex]10^{-11[/tex]N[tex](m/kg)^2[/tex]) * (6.0 x [tex]10^{24[/tex] kg) * (1.34 x [tex]10^{22[/tex]kg) / [tex](3.84 * 10^8 m)^2[/tex]
Calculating this expression, we find that the force of gravity between the Earth and the moon is approximately 1.982 x [tex]10^{20[/tex] Newtons.
b) Similarly, to calculate the force of gravity between the Earth and the sun, we can use the same formula:
m1 = 6.0 x [tex]10^{24[/tex] kg (mass of the Earth)
m2 = 2.00 x [tex]10^{30[/tex] kg (mass of the sun)
r = 1.49 x [tex]10^{11[/tex] m (distance between the Earth and the sun)
F = (6.67 x [tex]10^{-11} N(m/kg)^2[/tex]) * (6.0 x [tex]10^{24[/tex] kg) * (2.00 x [tex]10^{30[/tex] kg) / [tex](1.49 * 10^{11} m)^2[/tex]
Calculating this expression, we find that the force of gravity between the Earth and the sun is approximately 3.52 x [tex]10^{22[/tex] Newtons.
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A race car starts from rest and accelerates at 20.0 m/s2 for 5.0 s. It then moves with a uniform velocity for 60 s before it decelerates at 10.0 m/s2 until it comes to a stop. *Note: Show the complete solution by showing all of your work! (
a)Determine the velocity of the race car after 5.0 s.
(b)Determine the total distance traveled by the car between from the instant it started to move until it came to a stop.
(c)What was the total time that the car was in motion?
(a)The velocity of the race car after 5.0 s is 100.0 m/s. (b) The total distance traveled by the car from the instant it started to move until it came to a stop is 6250.0 m. (c) The total time that the car was in motion is 75.0 s.
Let's calculate the values step by step:
Given:
Acceleration during the first phase (a1) = 20.0 m/s²
Time during the first phase (t1) = 5.0 s
Uniform velocity phase (t2) = 60 s
Deceleration during the third phase (a3) = -10.0 m/s²
(a) To determine the velocity of the race car after 5.0 s, we can use the formula:
v = u + a × t
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
t is the time.
Since the race car starts from rest, the initial velocity (u) is 0 m/s.
v = 0 + (20.0 m/s²) × (5.0 s)
v = 100.0 m/s
Therefore, the velocity of the race car after 5.0 s is 100.0 m/s.
(b) To determine the total distance traveled by the car, we need to calculate the distance covered during each phase and sum them up.
Distance during the first phase:
Using the equation of motion:
s1 = u × t1 + (1/2) × a1 × t1²
Since the initial velocity (u) is 0 m/s:
s1 = (1/2) × (20.0 m/s²) × (5.0 s)²
s1 = 250.0 m
Distance during the second phase:
Since the car moves with a uniform velocity, the distance covered is:
s2 = v × t2
s2 = 100.0 m/s × 60 s
s2 = 6000.0 m
Distance during the third phase:
Using the equation of motion:
s3 = v × t3 + (1/2) ×a3 × t3
Since the final velocity (v) is 0 m/s:
s3 = (1/2) × (-10.0 m/s²) × t3²
The time during the third phase (t3) can be found by equating the final velocity to 0:
v = u + a × t
0 = 100.0 m/s + (-10.0 m/s²) × t3
t3 = 10.0 s
Substituting the value of t3:
s3 = (1/2) × (-10.0 m/s²) × (10.0 s)²
s3 = -500.0 m
Total distance traveled by the car:
Total distance = s1 + s2 + s3
= 250.0 m + 6000.0 m + (-500.0 m)
= 6250.0 m
Therefore, the total distance traveled by the car from the instant it started to move until it came to a stop is 6250.0 m.
(c) To determine the total time that the car was in motion, we add the
time for each phase:
Total time = t1 + t2 + t3
= 5.0 s + 60 s + 10.0 s
= 75.0 s
Therefore, the total time that the car was in motion is 75.0 s.
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which part of a centrifugal pump transmits energy in the form of velocity to the water? (299)
The impeller of a centrifugal pump is the component that transmits energy in the form of velocity to the water. It consists of curved blades or vanes that rotate, creating a centrifugal force that accelerates the fluid, increasing its velocity.
In a centrifugal pump, the impeller is responsible for transferring energy to the water in the form of velocity. The impeller is typically a wheel-like structure with curved blades or vanes.
When the pump is operational, the impeller rotates rapidly, drawing in water through the inlet. As the water enters the impeller, the curved blades exert a force on it, imparting angular momentum and causing it to move in a tangential direction.
Due to the centrifugal force generated by the rotating impeller, the water is propelled outward and accelerates as it moves away from the impeller's center.
This acceleration increases the water's velocity, transforming the potential energy into kinetic energy. The high-velocity water is then discharged from the impeller into the pump's volute or diffuser section, where the kinetic energy is gradually converted back into pressure energy.
The impeller is the crucial component of a centrifugal pump that transmits energy in the form of velocity to the water. Through its rotation and curved blades, it imparts angular momentum to the water, resulting in increased velocity and kinetic energy, which drives the flow of water through the pump system.
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The position of a particle is expression as = 2 + ^2 + ^3 , where r is in meters and t in seconds. a) Find the scalar tangential components of the acceleration at t =1s. b) Find the scalar normal components of the acceleration at t = 1s.
The angle between the velocity and acceleration vectors is given as;
cos(θ) = ([tex]v . a) / (∣v ∣ × ∣a ∣)v . a = 0 × 0 + 2 × 2 + 3 × 6 = 20So,cos(θ) = 20 / (√13 × √40)cos(θ) = 20 / 20cos(θ) = 1θ = cos^-1(1)θ = 0°[/tex]
The given position of a particle is,
`[tex]r = 2i + t^2j + t^3k`[/tex]
where r is in meters and t is in seconds. We have to find the scalar tangential components of the acceleration and scalar normal components of the acceleration at t = 1s.
The formula for the tangential component of acceleration is given as follows;
at = (v × a) / ∣v ∣
Where,
v = Velocity of the particle anda = Acceleration of the particle.
Using the above formula, we can find the scalar tangential component of acceleration at t = 1s.
Step 1: Velocity of the particle Velocity of the particle is obtained by differentiating the position of the particle with respect to time.
[tex]t = 1sv = dr / dtv = 0i + 2tj + 3t^2kv = 0i + 2j + 3k [put t = 1s]v = 2j +[/tex]
2: Acceleration of the particle Acceleration of the particle is obtained by differentiating the velocity of the particle with respect to time.
[tex]a = dv / dta = 0i + 2j + 6tk [put t = 1s]a = 0i + 2j + 6k[/tex]
So, the acceleration of the particle at
[tex]t = 1s is a = 0i + 2j + 6k.[/tex]
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Which of the following is not an advantage of nuclear power plants, when compared to fossil fuel plants? 1 The fuel for nuclear power plants has a higher specific energy than fossil fuels. 2 Nuclear power plants use renewable fuel. 3 Nuclear power plants do not produce greenhouse gases. 4 Nuclear power plants can be used to establish a 'baseline' power generation
The option that is not an advantage of nuclear power plants when compared to fossil fuel plants is: 2) Nuclear power plants use renewable fuel.
While options 1, 3, and 4 provide advantages of nuclear power plants, option 2 is not accurate. Nuclear power plants do not use renewable fuel. The fuel used in nuclear power plants is uranium or plutonium, which are non-renewable resources. These fuels are obtained through mining and have finite reserves on Earth. Once the fuel is used up, it cannot be replenished.
Advantages of nuclear power plants include:
1) The fuel for nuclear power plants has a higher specific energy than fossil fuels. This means that a small amount of nuclear fuel can produce a large amount of energy.
3) Nuclear power plants do not produce greenhouse gases. Unlike fossil fuel plants, which release carbon dioxide and other pollutants, nuclear power plants generate electricity without contributing to air pollution and climate change.
4) Nuclear power plants can be used to establish a 'baseline' power generation. Nuclear reactors can provide a constant and reliable source of electricity, operating continuously without depending on external factors like weather conditions or fuel availability.
Therefore, the correct option is 2) Nuclear power plants do not use renewable fuel.
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