The impulse shared by the object equals the difference in momentum of the object. In equation form,
F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.
How to calculate thrust provided by the rocket engines is 10 kN (10 000 N).?
a)There is this impulse-momentum change equation.
[tex]where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.[/tex]
We consider everything exits happen along a straight line, and gravitation does not participate.
So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.
[tex]$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$[/tex]
New velocity after engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.
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accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
I will run one mile in under 10 minutes 3 months from today.” is an example of a ___________ goal.
Long-term
Short-term
SMART
Both A and B
Answer:
long-term
Explanation:
usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.
how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m
2.89watts.
What is meant by sound intensity?The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:
Formula: [tex]I\frac{P}{4\pi r^{2} }[/tex]
I=1.6x10-3 w/m2
r=12m
[tex]I\frac{P}{4\pi r^{2} }[/tex]
[tex]P=I4\pi r^{2}[/tex]
[tex]=(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )[/tex]
The power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m [tex]=2.89watts.[/tex]
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explain the use of low energy x-ray?
Answer:
The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.
Which of the following is a force acting between objects that do not touch? Pls answer in 5-10 min!
Normal force
Frictional force
Electrical force
Applied force
Answer:
Electrical force
Explanation:
The electrons aren't actually touching each other. Normal is something on something pushed by gravity by they are touching, frictional is rubbing, applied is pushing
In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres is 46622.72N
Calculation and Parameters( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude and direction of the force on a
−7.35 C charge at this location.
Magnitude _____N
Direction?
- north
-south
-east
-west
Answer:
Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)
Explanation:
If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].
The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:
[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].
An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:
[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
What is meant by significant figures
Answer:
The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.
Answer:
each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.
Explanation:
three objects are located in a coordinate system as shown below in figure a.find the centre of mass. how does the answer change if the object on the left is displaced upward by 1.00m and the object on the right is displaced downward by 0.500m (b) the object as point particles
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
What is the center of mass of a system of particles?A place at which the entire mass of the body or all the masses of a system of particles appears to be concentrated is known as the center of mass of a body or system of particles. According to physics, the center of mass is a location where the total of the weighted relative positions of the distributed mass's points in space equals zero.a)
The center of mass of a three-particle system is expressed as
[tex]r_{cm}=[/tex] [tex]$$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$$[/tex]
When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )
[tex]$$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$$[/tex]
b)
When the two particles are shifted
[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]
Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
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A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
a) Calculate its angular velocity in rad/s .
Express your answer using three significant figures.
b)What is the linear speed of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
c)What is the acceleration of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
It is calculated that a) The angular velocity of the wheel is 272.13 rad/s,
b) On the edge of the grinding wheel, the linear speed is 47.62 m/s,
and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².
Calculation of angular velocity, linear speed & acceleration:
Provided that,
the diameter of the wheel = 0.35 m
So, the radius, r = 0.35/2 = 0.175 m
As 1 revolution = 2π rad
(a) the angular velocity, ω = 2600 rpm = [tex]\frac{2600 * 2\pi }{60}[/tex] rad/s
⇒ω = 272.13 rad/s
So, the angular velocity is 272.13 rad/s.
(b) The linear speed, v = r * ω
⇒v = 0.175 * 272.13
⇒v= 47.62 m/s
(c) The angular acceleration, [tex]a=\frac{v^{2} }{r}[/tex]
[tex]a = \frac{(47.62)^{2} }{0.175}[/tex]
⇒[tex]a[/tex] = 12958.08 m/s²
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If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
Answer:
The binding energy of sodium Na=5.407791×10⁹JExplanation:
Greetings !Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.
Formula : Eb=(Δm)c²where:Eb= binding energy .Δm= mass defect(kg) c= speed of light 3.00×10⁸ms¯¹Given valuesm= 18.02597c=3.00×10⁸ms¯¹required valueEb=?Solution:Eb=(Δm)c²Eb=(18.02597)*(3.00*10⁸ms¯¹Eb=5.407791*10⁹JIn transverse waves, there is no _________ motion.
A. Perpendicular
B. Vertical
C. Horizontal
D. Up and down
In a transverse wave, the disturbance are carried forward and the particles are undergoing to and fro motion. Hence in transverse waves, there is no horizontal motion.
A wave can be described as a disturbance that travels through a medium from one location to the other location without transporting any matter.
There are two types of matter,
(i) Transverse wave
(ii) Longitudinal wave
In a transverse wave, the particles of the medium move in a direction perpendicular to the direction of propagation of the wave. That means, the particles undergo to and fro motion about their mean position and moves in the direction of the wave propagation.
Examples are:
Radio wavesLight wavesWater waves when a stone is dropped in the pondWhereas longitudinal waves undergo back and forth motion with respect to the mean position.
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*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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a resistor wire of diameter 0.6cm has a resistivity of 1.0×00000010 ohms m. what length of the wire would be needed to make a 4 ohm resistor
The length of the wire is 1.12m
What is Resistance ?
Resistance is opposition to the current flowing in the circuit.
Given ,
D= 0.6m
R= 0.6/2m
R= 28Ω
rho = 1Ωm
R= rho ×l/A
where R is resistance , l is length ,A is area and rho is resistivity
By applying value and Solving we get,
l= 1.12m
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Please help, Class ends today!! will give brainiest if correct.
Fill in the blanks with agree or disagree
A list of planet-star radius ratios is an example of evidence.
I ____________ with this statement
Explain your answer, using an example from your investigation about habitable worlds.
A list of planet-star radius ratios is an example of evidence. I agree with this statement.
Why do i agree with the statement above?I agree with A list of planet-star radius ratios is an example of evidence because the Sample records that are taken for planet-star radius ratio are known to often gives a kind of an improved approximate of the stellar characteristics in regards to the smallest stars in the Kepler target list.
What does a planets radius mean?The planetary radius is known to be the distance that exist between a what we see as planet's center and its surface.
Therefore, one can say that planetary radius is known to be a measure of a given planet's size.
So, A list of planet-star radius ratios is a sample evidence and I do affirm the above statement.
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why does a bus look red in the daylight?
Answer:
The red color in the sky at sunset (and sunrise) is due to an effect called Rayleigh scattering.
Explanation:
Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
A. it gets stuck in the atmosphere
B. it is magnetically attracted
C. it is attached by a rope to the Earth
D. it falls at the same rate the Earth curves
The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.
Newton's Cannonball:Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.
Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.
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Three bees are oriented as shown in the figure. B1 has +17 µC of charge, B2 has −5 µC of charge, and B3 has +26 µC of charge.
B1 to b2 5 cm
B2 to b3 10 com
Bee number 3 is stationed at the observation location for this problem, and we want to find the net electric field at B3. We'll do this in a few steps.
(a) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B1? Remember that the components can be positive or negative depending on their directions along the x or y axis.
E1x =
E1y =
(b) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B2? Again, remember that the components can be positive or negative depending on their directions along the x or y axis.
E2x = N/C
E2y =
(c) What are the magnitude and direction of the net electric field at the observation location where B3 is resting?
magnitude
N/C
direction
° counterclockwise from the +x-axis
(d) What is the magnitude of the force on B3 due to this net electric field?
Given the following data:
Charge of B₁ = +17 µC to C = 17 × 10⁻⁶ C.Charge of B₂ = -5 µC to C = -5 × 10⁻⁶ C.Charge of B₃ = +26 µC to C = 26 × 10⁻⁶ C.Radius of B₁ to B₂ = 5 cm to m = 0.05 meter.Radius of B₂ to B₃ = 10 cm to m = 0.1 meter.How to determine the x and y components of the electric field?First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.
The electric field due to B₁ to B₃ is given by:
E₁ = kq₁/r₁²
E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)
E₁ = 153 × 10³/0.0125
Electric field, E₁ = 12.24 × 10⁶ N/C.
The electric field due to B₂ to B₃ is given by:
E₂ = kq₂/r₂²
E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)
E₂ = 45 × 10³/0.01
Electric field, E₂ = 4.5 × 10⁶ N/C.
Also, the magnitude of the angle formed is given by tan trigonometry:
Tanθ = 5/10
Tanθ = 0.5
θ = tan⁻¹(0.5)
θ = 26.56°.
Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁x = E₁cosθ - E₂
E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶
E₁x = 10.95 × 10⁶ - 4.5 × 10⁶
E₁x = 6.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁y = -E₁sinθ
E₁y = -12.24 × 10⁶ × sin(26.56)
E₁y = -12.24 × 10⁶ × 0.4471
E₁y = -5.5 × 10⁶ N/C.
Also, the magnitude of the electric field is given by:
Exy = √(E₁x² + E₁y²)
Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)
Exy = √4.225 × 10¹³ + 3.025 × 10¹³)
Exy = √(7.25 × 10¹³)
Exy = 8.5 × 10⁶ N/C.
Part B.We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:
E₂x = -E₂sinθ
E₂x = -4.5 × 10⁶ × sin(90)
E₂x = -4.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₂y = E₂cosθ
E₂y = -4.5 × 10⁶ × cos(90)
E₂y = 0 N/C.
How to calculate the net electric field at the observation location?The magnitude of the net electric field at B₃ is given by:
Eₙ = √(E₁x + E₂x)² + E₁y²)
Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)
Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)
Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)
Eₙ = √(3.65 × 10¹³)
Eₙ = 6.04 × 10⁶ N/C.
For the direction, we have:
Tanθ = x/y
Tanθ = 6.5/-5.5
Tanθ = -1.1818
θ = tan⁻¹(-1.1818)
θ = 49.76°.
Part C.The magnitude of the force on B₃ due to this net electric field is given by:
F = B₃ × Eₙ
F = 26 × 10⁻⁶ × 6.04 × 10⁶
F = 157.04 Newton.
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A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate of 0.21 m/s^2 and catches the ball after 3.8s. What was the dogs velocity when it caught the ball? *Assume here that running to the right is positive and running to the left is negative.
A- 7.96
B- 10.569
C- 6.782
D- 4.798
E- none
Answer:
D.-4.798m/s
Explanation:
Greetings !
Given values
[tex]u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec[/tex]
Solve for V of the given expression
Firstly, recall the velocity-time equation
[tex]v = u + at[/tex]
plug in known values to the equation
[tex]v = (4) + (0.21)(3.8)[/tex]
solve for final velocity
[tex]v = 4.792ms[/tex]
Hope it helps!
A 29-g rifle bullet traveling 200 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.5-m-long string, which makes the pendulum swing upward in an arc. (Figure 1)
a) Determine the vertical component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
b)Determine the horizontal component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.
What is the velocity of the pendulum just after the collision?As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity
=> 0.029kg × 200 = 3.429 × velocity of pendulum
=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s
What is the amplitude of the pendulum's motion?As per conversation of energy,1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²
Here, k = mg/L = (3.429×9.8)/2.5 = 13.44 N/mSo, 3.429×1.7²= 13.44× amplitude²=> 10 = 13.44× amplitude²
=> Amplitude = √(10/13.44)
= 0.86m
What is the angle made by the pendulum with vertical direction at largest displacement?Angle = tan inverse (amplitude/length of pendulum)
= tan inverse (0.86/2.5)
= 20°
What is the maximum displacement along vertical direction?Maximum displacement along vertical direction= L - L×cos20°
= 2.5 - 2.5×cos20°
= 0.15 m
What's the maximum displacement along horizontal direction?Maximum displacement along horizontal direction= Lsin20°
= 2.5 × sin 20°
= 0.85m
Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.
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The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.
(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
Increase in the sound levelThe INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;
Increase in sound level = final sound level - original sound level
Increase in sound level = 86 dB - 85 dB = 1 dB
Factor of sound level increasefrom the graph at 1.5 hours/day, sound level = 97 dB
Increase in sound intensity = final sound level - original sound level
Increase in sound intensity = 116 dB - 97 dB = 19 dB
Factor increase = 19/97 = 0.196 = 19.6%
Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
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What is a small portable device that counts every step taken throughout the day?
A.stethoscope
B.pedometer
C.otoscope
D.thermometer
Answer: B
Explanation: i learned it last year
Pedometer is a small portable device that counts every step taken throughout the day. The correct option is (B).
A pedometer is a small portable device that counts the number of steps taken by an individual throughout the day. It is typically worn on the waist or carried in a pocket and uses a mechanism to detect body motion and count each step.
Pedometers are commonly used for tracking physical activity and encouraging individuals to achieve daily step goals as part of their fitness routines.
Therefore, Pedometer is a small portable device that counts every step taken throughout the day. Choose the option (B).
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A light bulb in a camper’s flashlight is labeled 4.1 V, 0.40 A. Find the equivalent current if three of these light bulbs are connected in parallel to a standard C size 1.5 V battery.
Explanation:
first rule of parallel circuits : the voltage is the same in the whole circuit, no matter where we measure.
so, we have 1.5V at every bulb.
second rule of parallel circuits : the total current is the sum of the individual branch currents.
so, if the battery really allows it, then we have 3 times 0.4 A = 3×0.4 = 1.2 A as current in the circuit.
Suppose that the math man, a super here that fights crime with math and physics, can decelerate the rate of gravity. During a recent fight with a diamond thief, Math man fell from the top of a 500 meter building. The equation D=t^2*.25. How long us he falling from the top of the building to the ground?
Answer:
44.7 seconds
Explanation:
D = 500 m
500 = .25 t^2
500/.25 = t^2
2000 = t^2
t = 44.7 seconds
formula for conductivity
Answer:
Thermal conductivity formula
k is the thermal conductivity (Wm -1 K -1)Q is the amount of heat transferred through the material (Js -1)A is the area of the body (m 2)ΔT is the temperature difference (K)Explanation:
Please help 25 points
Asteroids X, Y, and Z have equal mass of 3.0 kg each. They orbit around a planet with M = 7.20×10^24 kg. The orbits are in the plane of the paper and are drawn to scale.
The three asteroids orbit in the same counterclockwise direction.
The angular velocity of X at i is .... that of Y at i.
The angular velocity of Y at c is .... that at i.
The angular momentum of Z at c is .... that at v.
The angular velocity of X at i is .... that at r.
The angular momentum of Y is .... that of X.
The period of Z is .... that of X.
The period of Y is .... that of X.
(Options are: greater than, less than, equal to)
This is a series of analysis of Angular velocity as relates to an asteroid's orbit. See the explanation below.
What is an asteroid?Asteroids are stony bodies that circle the Sun.
Although asteroids circle the Sun in the same way as planets do, they are far smaller.
Hence, from the information given:
A) The square of the period is proportional to the cube of the semimajor axis, according to Kepler's third law. As a result, the period of Y equals (E) the period of Z.
B) Angular momentum is preserved here, hence it is equivalent (E).
C) As eccentricity increases, so does the angular momentum. In this case, Y and Z have the same period, and both satellites cover the same proportion of the territory in the same length of time.
This indicates that a satellite on Z must cover a lesser area in a given period of time than a satellite on Y. The area swept is approximately 1/2 the radius times the tangential displacement.
Because both satellites have the same "radius" at point y, the satellite on Z must have a lower tangential velocity than the one on Y. As a result, Y has more angular momentum than (G) Z.
D) Using Kepler's third law, X's period is bigger than (G) Z's.
E) In a circular orbit, the angular velocity is constant. As a result, the angular velocity of Y at y equals (E) that at s.
F) Z's angular velocity at c is smaller than (L) at i.
G) Y's angular velocity at y is larger than (G) Z's at y.
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The y-position of a damped oscillator as a function of time is shown in the figure.
This function can be described by the y(t) = [tex]A_{0}[/tex][tex]e^{-btx}[/tex]cos(ωt) formula, where [tex]A_{0}[/tex] is the initial amplitude, b is the damping coefficient and ω is the angular frequency.
1. What is the period of the oscillator? Please, notice that the function goes through a grid intersection point.
2. Determine the damping coefficient.
(1) The period of the oscillator is 1 second.
(2) The damping coefficient is 0.93.
What is period of oscillation?
The period of oscillation is the time taken to make one complete cycle.
From the graph, the time taken to make one complete oscillation is 1 second.
Damping coefficientequation of the wave is given as;
y(t) = Ae^(-btx) cos(ωt)
at time, t = 0, y = 3.53.5 = Ae^(-0) cos(0)
3.5 = A x 1
A = 3.5 cm
at time, t = 1 cm, y = - 3cm-3 = 3.5e^(-bx) cos(ω)
-3/3.5 = e^(-bx) cos(ω)
-0.857 = e^(-bx) cos(ω)
-0.857 / cos(ω) = e^(-bx)
ln[-0.857 / cos(ω)] = -bx
ln[-0.857 / cos(ω)] / b = - x ---- (1)
at time, t = 2 cm, y = - 2cm-2 = 3.5e^(-2bx) cos(2ω)
-0.57 = e^(-2bx) cos(2ω)
ln[-0.57 / cos(2ω)] = -2bx
ln[-0.57 / cos(2ω)] /2b = - x ------(2)
solve (1) and (2)
ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b
-0.57 / cos(ω) = 2(-0.857 / cos(ω))
2(-0.857/cosω) = -0.57/cos2ω
-(2 x 0.857) / (-0.57) = cosω/cos 2ω
3 = cosω/cos 2ω
3(cos 2ω) = cosω
3(2cos²ω - 1) = cos ω
6cos²ω - 6 = cosω
6cos²ω - cosω - 6 = 0
let cosω = y
6y² - y - 6 = 0
solve the quadratic equation;
y = 1.1 or -0.92
cosω = -0.92
ω = arc cos(-0.92)
ω = 2.74 rad/s
From equation (1)
ln[-0.857 / cos(ω)] / x = -b ---- (1)
let x = 1
ln(-0.857/cos(2.74) = -b
-0.93 = -b
b = 0.93
Thus, the damping coefficient is 0.93.
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Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
19.5 km/s
27.5 km/s
11.2 km/s
20.5 km/s
The orbital speed of an ice cube in the rings of Saturn is 355358.97m/s
Law of gravitationAccording to the gravitation law, the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. Mathematically;
F = GMm/r²
where
m = mass of ice cube and
s = Gm1/r^2
Hence,
F = sm2
On rearranging,
s = m2/F
let V = orbital speed
centripetal acceleration = V^2/r
Such that;
V²/r = Gm/r²
V² = Gm/r
V = √Gm/r
Substitute the given parameters
V = √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5
V = 355358.97m/s
Hence the orbital speed of an ice cube in the rings of Saturn is 355358.97m/s
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PLS HELP 20 PIONTS!!
It is critical for scientists to be able to describe components of a system quantitatively. Explain why it is important to be able to describe a system quantitatively, using an example from your investigation about habitable worlds
A quantitative description of a system is important as it provides information about the carrying capacity of habitable worlds.
What is a quantitative description of the components of a system?Quantitative description refers to the description of a system which is focused on the numerical value of the properties or components of the system.
For example, a quantitative description of the components of a given habitat will be focused on the number of the individual species in the habitat. It will also be focused on the numerical value of the non-living components of the system and how such values affect the living components of the system numerically.
This information will then be used by scientists to see how modifications of the various quantitative components of the habitat will help to improve the chances of survival of species found in the habitat. This leads to such concepts as the carrying capacity of a habitat which is the maximum number of species that the available resources is a habitat can easily sustain.
In conclusion, a quantitative description is important to in investigations about the habitable world.
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A child in a boat throws a 5.90-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 25.0 kg and the mass of the boat is 38.4 kg. (Figure 1)
Calculate the velocity of the boat immediately after, assuming it was initially at rest.
Express your answer to three significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the box and negative value if the direction of the velocity is in the direction opposite to the velocity of the box.
-0.930 m/s is the velocity of the boat.
Given:Mass of child and boat , [tex]m_1[/tex] = (25.0 + 38.4 )kg
= 63.4 kg
Mass of the package, [tex]m_2[/tex] = 5.90 kg
Velocity of package thrown from boat , [tex]v_2[/tex] = 10.0m/s
[tex]v_1 =?[/tex]
Initial velocity v = 0 m/s
As the boat is at rest, [tex](m_1 + m_2) v=0[/tex]
According, to the law of conversation of momentum;
∴ Momentum before = Momentum after
[tex]( m_1 + m_2 ) v = m_1v_1 + m_2v_2\\0 = m_1v_1 + m_2v_2\\0 = 63.4 v_1 + 5.90*10.0\\63.4 v_1 = - 5.9\\v_1 = - 0.930 m/s[/tex]
Negative direction shows the velocity in the direction opposite to the motion of the package.
Therefore, -0.930 m/s is the velocity of the boat.
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