(a) The amplitude of the magnetic field produced is 2.667 mV/m.
(b) The intensity of the laser is 2.832 W/m².
(c) The power delivered by the laser is 2.22 x 10⁻⁶ W.
Amplitude of the magnetic field producedIn electromagnetic waves, the amplitude of magnetic field is the maximum field strength of the magnetic fields.
B₀ = E₀/c
where;
E₀ is the amplitude electric fieldB₀ is the amplitude magnetic fieldc is speed of lightB₀ = (0.8 MV/m) / (3 x 10⁸)
B₀ = (0.8 x 10⁶ V/m) / (3 x 10⁸)
B₀ = 2.667 x 10⁻³ V/m
B₀ = 2.667 mV/m
Intensity of laserThe intensity of the laser is calculated as follows;
I = ¹/₂ε₀E₀²
I = (0.5)(8.85 x 10⁻¹²)(0.8 x 10⁶)²
I = 2.832 W/m²
power delivered by the laserP = IA
where;
A is the area of the beamA = πd²/4
where;
d is diameterA = π(1 x 10⁻³)²/4
A = 7.854 x 10⁻⁷ m²
Power = (2.832 W/m²) x (7.854 x 10⁻⁷ m²)
Power = 2.22 x 10⁻⁶ W
Thus, the amplitude of the magnetic field produced is 2.667 mV/m.
The intensity of the laser is 2.832 W/m².
The power delivered by the laser is 2.22 x 10⁻⁶ W.
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The object that ends up with less electrons has a __________ charge.
A. Positive
B. Not enough info
C. Neutral
D. Negative
Reset Selection
The object that ends up with less electrons has a positive charge; option A.
What are electrons?Electrons are negatively charged particles which form a part of the three fundamental particles in an atom.
Electrons can easily be removed from the vicinity of atoms making the atom to become positively charged. On the other hand, when atoms gain extra electrons, they become negatively-charged.
On way of charging objects is by rubbing them against other objects in order to gain extra electrons or to lose electrons.
When two objects are rubbed together, the object that ends up with less electrons has a positive charge, whereas the object that ends up with more electrons has a negative charge.
In conclusion, objects can become charged when they are rubbed against each other, and they can either become negatively or positively-charged by gain or loss of electrons.
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The object indicated here that ends up with fewer electrons has an unknown charge (not enough info, Option B).
What does atomic charge mean?The atomic charge makes reference to the overall charge of an atom, which can be evidenced by knowing its net charge (positive or negative charge).
The net charge of a molecule and/or atom is obtained by calculating the amount of electrically negative electrons (e-) and the number of positive protons.
Neutrons are another type of subatomic particle that have not net charge (neither positive nor negative).
In conclusion, the object indicated here that ends up with fewer electrons has an unknown charge (not enough info, Option B).
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Maya made this picture to represent a chemical reaction:
Which of the following statements best explains the type of chemical reaction represented by Maya's picture?
A. It is neither a synthesis reaction nor a decomposition reaction because the total mass of the products is less than the total mass of the reactants.
B. It is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.
C.It represents a synthesis reaction because the same atoms are present in the reactants and products.
D. It represents a decomposition reaction because two reactants break apart and form two products.
The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.
What is a chemical reaction?A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.
A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.
A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.
Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.
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which element has more than 5 valency
Answer:
Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.
Explanation:
The electrons found in an element's outermost atomic shell are known as valence electrons.
Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.
However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.
Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.
Thank you ,
Eddie
*please refer to photo*
Use meters and seconds for all calculations. Show the appropriate units for measured and calculated values
Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.
What is average velocity?An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.
What is an average?An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.
How to calculate the average velocity?Mathematically, the average velocity for this data set would be calculated by using this formula:
Average = [F(v)]/n
Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5
Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:
v₁ = 100 m/sv₂ = 150 m/sv₃ = 200 m/sv₄ = 250 m/sv₅ = 300 m/sSubstituting the parameters into the formula, we have:
Vavg = [300 + 450 + 500 + 250 + 300)/5
Vavg = 1800/5
Vavg = 360 m/s.
Next, we would calculate the percent difference by using this formula:
[tex]Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100[/tex]
Percent difference = [360 - 343]/360 × 100
Percent difference = 17/360 × 100
Percent difference = 0.0472 × 100
Percent difference = 4.72%.
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(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C
If a charge of −4.72 µC is placed at this point, what are the magnitude and direction of the force on it? Magnitude _______N
Direction?
- toward the left
- upward
-downward
- toward the right
The electric field strength at a point 1.00 cm to the left of the middle is -2.0 x 10⁷ N/C.
The magnitude of the force is 94.4 N and direction of the force on it towards the right.
Electric field strengthThe electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first charge
E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second charge
E2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third charge
E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
Net electric fieldE = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C
E = -2.0 x 10⁷ N/C
Force on the charge −4.72 µCF = Eq
F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = 94.4 N
Thus, the direction of the force will be towards the right.
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NASA wants a satellite to revolve around Earth 3 times a day. What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg)
Answer:
Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].
Explanation:
Assume that the radius of this orbit is [tex]r[/tex].
Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].
The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].
The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].
Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].
In other words:
[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].
[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].
The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:
[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].
Substitute this value into the expression for [tex]r[/tex] and evaluate:
[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].
(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)
How is faraday's law applicable in Electronic Drum
Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum
What is Faraday's law?Faraday's law states that a changing magnetic field through an area or equivalently, a changing area with constant field will cause a voltage.
So therefore, Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum
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A visitor at a Michigan lighthouse is trying to measure the height of the lighthouse. She has a big spool of string, but she doesn't have a measuring tape. She has a wrist watch with a timer function on it. She attaches a weight to the end of the string and she makes a simple pendulum. She hangs the pendulum down from the top of the spiral staircase and she measures the period of the pendulum. The period of the pendulum is 8.48 s. What is the height of the light house?
The height of the light house is 17.86m.
To find the answer, we have to know about the simple pendulum.
How to find the height of the light house?We have the expression for time period of the pendulum as,[tex]T=2\pi \sqrt{\frac{l}{g} } \\[/tex]
where, g is acceleration due to gravity, and l is the length of the pendulum.
For the pendulum with time period T kept at some height under the influence of gravity, then the height will be equal to,[tex]h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m[/tex]
Thus, we can conclude that, the height of the light house is 17.86m.
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According to current evidence, what happens when gays or lesbians rear a child?
a. the child is more likely than average to develop depression or anxiety disorders.
O b. the child develops about the same as other children.
O c. the child is more likely than average to become gay or lesbian.
O d. the child does poorly in school.
Answer: The answer is A.
Explanation:
Just because u have gay parents doesnt mean theyre bad lol XD
The velocity of a body is given by the equation v= a + bx, where 'x' is displacement. The unit of b is .......
Answer:
s^ -1 ( or 1/sec)
Explanation:
Velocity is given in units of displacement / sec
like feet /sec or m/sec
so b would have units of s^-1
(or perhaps a more general term would be time^-1)
Is India a rich country?
Explanation:
India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.
Answer:
50505050128128128128
Explanation:
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
a)What is the velocity of the space capsule after the push in the reference frame?
Express your answer to two 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 astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut.
b)If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other?
Express your answer to three significant figures and include the appropriate units.
c)What is the kinetic energy of the astronaut after the push in the reference frame?
Express your answer to three significant figures and include the appropriate units.
d)What is the kinetic energy of the capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units.
The change in the speed of the space capsule will be -0.189 m/s.
The average force exerted by each on the other will be 567 N.
The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
Given:Mass of the astronaut, [tex]m_a[/tex] = 126 kg
Speed he acquires, [tex]v_{a}[/tex] = 2.70 m/s
Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg
The initial momentum of the astronaut-capsule system is zero due to rest.
[tex]P_f = m_av_a + m_cv_c[/tex]
[tex]P_I[/tex] = 0
[tex]m_av_a + m_cv_c = 0[/tex]
[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]
[tex]= \frac{126* 2.70}{1800}[/tex]
[tex]= - 0.189[/tex] m/s
Therefore,
According, to the impulse-momentum theorem;
FΔt = ΔP
ΔP = m Δv
ΔP = 126×2.70
= 340.2 kgm/sec
t is time interval = 0.600s
F = ΔP/Δt
F = 340.2/0.600
= 567 N
Therefore, the average force exerted by each on the other will be 567 N.
The Kinetic Energy of the astronaut;
K.E = [tex]\frac{1}{2} m v^2[/tex]
[tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]
= 459.27 J
The Kinetic Energy of the capsule;
K.E = [tex]\frac{1}{2} m v^2[/tex]
= [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]
= 32.14 J
Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
Answer:
Kinetic energy of the projectile at the vertex of the trajectory: [tex]900\; {\rm J}[/tex].
Work done when firing this projectile: [tex]2500\; {\rm J}[/tex].
Explanation:
Since the drag on this projectile is negligible, the horizontal velocity [tex]v_{x}[/tex] of this projectile would stay the same (at [tex]30\; {\rm m\cdot s^{-1}}[/tex]) throughout the flight.
The vertical velocity [tex]v_{y}[/tex] of this projectile would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] at the vertex (highest point) of its trajectory. (Otherwise, if [tex]v_{y} > 0[/tex], this projectile would continue moving up and reach an even higher point. If [tex]v_{y} < 0[/tex], the projectile would be moving downwards, meaning that its previous location was higher than the current one.)
Overall, the velocity of this projectile would be [tex]v = 30\; {\rm m\cdot s^{-1}}\![/tex] when it is at the top of the trajectory. The kinetic energy [tex]\text{KE}[/tex] of this projectile (mass [tex]m = 2.0\; {\rm kg}[/tex]) at the vertex of its trajectory would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}[/tex].
Apply the Pythagorean Theorem to find the initial speed of this projectile:
[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the initial kinetic energy [tex]\text{KE}[/tex] of this projectile would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}[/tex].
All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be [tex]2500\; {\rm J}[/tex].
Nina is doing a workout where she runs at 50% speed for 3 minutes, 75% speed for 2 minutes, and 100% speed for 30 seconds. This is an example of
a. Specificity training
b. Overload training
c. Circuit training
d. Interval training
What is the equation for the enthalpy of vaporization?
A. Q = m*Lv
B. Q = m/Lv
C. Q = 1/m*Lv
D. Q = Lv/m
Answer:
A: Q = m*Lv
Explanation:
The equation formula for the enthalpy of vaporization is;
Q = m*ΔHv
Where;
q = heat energy
m = mass
ΔHv = heat of vaporization
How much energy is needed to move an electron in a hydrogen atom from the ground state (n = 1) to n = 3?
The energy needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3 will be 1.93 *10^-18J and 12.09 eV.
How to compute the value?The following can be deduced:
Energy of electron in hydrogen atom is
En = -13.6 /n2 eV
where n is principal quantum number of orbit.
Energy of electron in first orbit = E1 = -13.6 / 12 = - 13.6eV
Energy of electron in third orbit = E3 = -13.6 /32 = -1.51 eV
Energy required to move an electron fromfirst to thirdorbit ΔE = E3- E1
ΔE = -1.51 - ( 13.6) = 12.09 eV
Energy in Joule = 12.09 *l/× 1.6 × 10^-19 = 1.93 × 10^-18 J.
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Complete question:
How much energy is needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3? Give theanswer (a) in joules and (b) in eV.
An object with an initial velocity of
0.12rad accelerates at 0.11rad over a
distance of 0.25 radians. What is the
final angular velocity of the object?
rad
S
Answer:
100rad because it angular velocity
a 30kg crate is pulled up a ramp 15m long and 2m high by a constant force of 100n. the crate starts from resta nd has a velocity of 2m/s when it reaches the ramp. what is the frictional force between the crate and the ramp. use the principal of conservation energy.
The frictional force between the crate and the ramp is determined as 35.2 N.
Energy lost to frictional forceApply the principle of conservation energy to calculate the change in the energy of the crate.
Change in energy of the crate = energy loss to friction
P.Ei - K.Ef = E
mgh - ¹/₂mv² = E
where;
m is mass of the crateh is vertical height traveled by the cratev is the final velocity of the crate(30)(9.8)(2) - (0.5)(30)(2²) = E
528 J = E
Frictional force between the crate and the rampE = Fd
where;
F is the frictional forced is the distance traveled by the crateF = E/d
F = (528)/(15)
F = 35.2 N
Thus, the frictional force between the crate and the ramp is determined as 35.2 N.
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(refer to photos attached. Example of previous question with wrong/correct answers example, and current question needing to be solved)
Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C
If a charge of −3.94 µC is placed at this point, what are the magnitude and direction of the force on it?
Magnitude _______N
Direction?
- toward the left
- upward
-downward
- toward the right
(a) The electric field strength at a point 1.00 cm to the left of the middle is 2.0 x 10⁷ N/C.
(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.
Electric field strengthThe electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first chargeE1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second chargeE2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third chargeE3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
Net electric fieldE = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)
E = +2.0 x 10⁷ N/C
Force on the charge −4.72 µCF = Eq
F = 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = -94.4 N
Thus, the direction of the force will be towards the left.
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iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weight of iron ball in air?
Answer:
453 gm
Explanation:
Immersed objects are buoyed up by force equal to mass of displaced liquid
400 + 53 = 453 gm in air
Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.3 min to turn through 22 complete revolutions before reaching its final speed.
a)What was its angular acceleration (assumed constant)?
Express your answer using two significant figures.
b)What was its final angular speed in rpm ?
Express your answer using two significant figures
(a) The angular acceleration will be 26.035 rev/[tex]min^{2}[/tex].
(b) The final angular velocity is expected to be 33.846 rev/min.
Given.
t=1.3 min, Θ=22 rev, [tex]ω_{i}[/tex]=0
We know, Θ= [tex]ω_{i}[/tex]t+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]t^{2}[/tex]
22=0+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]1.3^{2}[/tex]
[tex]\alpha[/tex]=26.035 rev/[tex]min^{2}[/tex]
[tex]ω_{f} =ω_{i}+\alpha t[/tex]=0+26.035*1.3=33.846 rev/min
Angular velocityAn object's rate of change in angular position or orientation over time is depicted by its angular velocity, rotational velocity, or both ( or ), also known as the angular frequency vector (i.e. how quickly an object rotates or revolves relative to a point or axis). The direction of the pseudovector is normal to the instantaneous plane of rotation or angular displacement, and its magnitude denotes the angular speed, or the rate at which the item rotates or revolves. It is customary to use the right-hand rule to specify the direction of angular motion. A general definition of angular velocity is "angle per unit time" (angle replacing distance from linear velocity with time in common). Radians per second is how angles are measured in the SI.
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The cricket ball has a mass of 0.16kg and it hits the bat with a speed of 25 m/s. After being in contact with the bat for 0.0013 s, the ball rebounds with a speed of 22 m/s in the direction exactly opposite to its original direction.
A) state the difference between speed and velocity
B) calculate.
I) the change in velocity of the cricket ball
Ii) the average acceleration of the ball whilst it is in contact with tge bat
Iii) the average force exerted on the ball by the bat
change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².
Give some difference between velocity and speed.1) Speed is scalar quantity but velocity is vector quantity.
2) Speed is distance followed by an individual with respect to time and velocity is displacement with respect to time .
1) final velocity- intial velocity
25-22= 3m/s
2) The force is
F= ma
. 0.16×30= 4.8N
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A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N. What is the speed of transverse waves on the wire?
The speed of transverse waves on the wire is 1.555.
Given: Length= 1.150 m
Mass= 4.80 g
Speed =mass/length
U=M/L formula
U=1.150 *10^(-3)/4.80
v=√t/u
v=√580*4.80/1.150*10^(-3)=1.555
1.555
The derivative of the displacement with respect to time gives the transverse velocity in the y direction, as velocity is the rate of change of position: Where k = 2/ is referred to as the wave number and = 2f as before, v(x,t) = y(x,t)/t = -A cos (k x - t + ).The linear density and tension v=FT can be used to determine the wave's speed. v = F T μ . According to the equation v=FT, v = F T, the tension would need to be increased by a factor of 20 if the linear density were to be doubled by almost as much.
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A fighter plane is flying overhead at mach 1.20. What angle does the wave front of the shock wave produced make relative to the plane's direction of motion (in degrees)?
Answer: 56.44°
Explanation:
Given:
Let u represent the current speed of the plane, 1.2 MachConverting to SI Units (m/s):
= (1.2 mach)(340 ms^-1 / 1 Mach)
u = 408 m/s
Speed of sound in air, v = 340 m/sFind:
Angle the wave front of the shock wave relative to the plane's direction of motion, θWe have, sinθ = speed of sound / speed of object
sinθ = v / u
θ = sin^-1 (v / u)
= sin^-1 (340 / 408)
θ = 56.44°
Physics occurs all the time but often goes unnoticed. Here is your chance to reflect on physics in action. Other than the examples used in this lesson, think of a time where you witnessed the conservation of angular momentum. Describe the objects that had angular momentum and how angular momentum was conserved. You may also create an example if you cannot recall one in your personal experience.
Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
What is principle of conservation of angular momentum?The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no external torque acting on the object.
Angular momentum of a system is conserved as long as there is no net external torque acting on the system.
Examples of conservation of angular momentumthe spinning ice skatersomeone spinning in an office chaira child spinning on roller coasterThus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
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You are designing a delivery ramp for crates containing exercise equipment. The 1390 N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0∘. The ramp exerts a 515 N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp.
The maximum force constant of the spring Kmax is 2337.9 N/m.
What is force constant of a spring?The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.
Force constant is denoted by K and its unit is N/m.
Force = K * xWhere;
K = spring constant
x = displacement
The work done by the spring is given below as follows:
Work done = Fx/2
Kinetic Energy = mv²/2
Force on an inclined plane = mgsinθ
Total force, F = mgsinθ + frictional force
F = 1390 * sin 22° + 515
F = 1035.7 N
Work done = change in KE
Fx/2 = mv²/2
Fx = mv²
m = 1390/9.81 = 141.692
Solving for x;
x = mv²/F
x = 141.692 * 1.8²/1035.7
x = 0.443 m
The maximum force constant of the spring Kmax = 1035.7/0.443
Kmax = 2337.9 N/m
In conclusion, the maximum force constant of the spring is the ratio of the total force and displacement.
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Note that the complete question is given below:
You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria
A jet of water squirts out horizontally from a hole near the bottom of the tank, as seen in the figure.
If the hole has a diameter of 3.13 mm, what is the height, h, of the water level in the tank? Assume that x = 1.33 m and y = 1.72 m.
The height of the water level in the water tank is 0.011 m.
What is the time taken by a fluid particle to reach the ground?From Newtown's equation of motion, y=U×t+(1/2)gt²As, initial velocity (U) is zero, t= 2y/gt = (2× 1.72)/9.8 = 0.35 s
What's the velocity along x direction?Velocity along x direction= x × t
= 1.33 × 0.35
= 0.46 m/s
What's the height of water level in the tank?As per Bernoulli's theorem, 1/2 × ρ × v² = ρ × g× h
=> h = v²/2g = 0.46²/(2×9.8)
= 0.011 m
Thus, we can conclude that the height of water level is 0.011m.
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An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How will the final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected?.
The kinetic energy of the ejected electron (photoelectron) is equal to the energy of the photon minus the work function (E required to eject the photoelectron).
The process of final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected -
The electron has a lot of potential energy and very little kinetic energy when it is liberated from the cathode (as it is not moving yet or moving very slowly). We are currently on the left side of the picture above. The electron accelerates and speeds up as it goes away from the cathode and toward the anode. As a result, the kinetic energy rises.The electron is likely to have the highest kinetic energy and the lowest potential energy when it reaches the anode.Therefore, the kinetic energy at the end of the electron's journey is approximately equal as the potential energy at the beginning (when the electron was released).Learn more about kinetic energy
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In the hydrogen atom, which electronic transition corresponds to the largest energy emission?
In hydrogen atom the electronic transition from n= ∞ to n=1 corresponds to largest energy emission.
What is electronic transition ?Electronic transition is the jump of an electron from one energy level to another energy level .
Which electronic transition in Hydrogen atom corresponds to highest energy emission?The fromula for energy emission isE = hc/λ
where E= emited energyh= plank constant =6.62*10^(-34)
c=speed of light
λ= wavelength of emited electron .
The electronic transition from infinity to ground state corresponds to lowest wavelength .Thus , we can conclude that the electronic transition from infinity to n=1 (ground state) corresponds to largest energy emission .
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A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?
A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then
(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s
(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.
c) The speed of the 0.950- kg puck after the collision is 2.82 m/s
d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.
Given:Mass of ice puck, m₁ = 0.350 kg
Mass of another puck, m₂ = 0.950 kg
Velocity of ice puck, v₁ = 5.22 m/s
Velocity of another puck, v₂ = 0 m/s
[tex]v^{'}_1= ?[/tex]
[tex]v^{'}_2= ?[/tex]
[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]
[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]
[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]
Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.
[tex]v^{'}_2 = v^{'} + v_1[/tex]
[tex]= -2.40+5.22[/tex]
[tex]= 2.82 m/s[/tex]
Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.
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