While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.
Answers
Answer:
a) W = 25.872 J
b) - 35.28 J
c) - 9.408
Explanation:
a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)
F = - mg (gravity is acting downwards)
F = - 0.6 × 9.8 = - 5.88 N
(H₂ - H₁) = (1.6 - 6) = - 4.4 m
W = (-5.88)(-4.4) = 25.872 J
b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J
c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J
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What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.
Measure the circumference of the tire before and after riding.
B.
Measure the total distance traveled on his bike and divide this by how long it took him.
C.
Measure the wear on his treads before and after riding a certain number of laps.
D.
Time how long it takes him to ride 5 laps around his cul-de-sac.
Answers
Answer:
C.
Measure the wear on his treads before and after riding a certain number of laps.
Answer:
Measure the wear on his treads before and after riding a certain number of laps.
Explanation:
By riding in a circular motion the inside of the tire will be in contact with the road more than the outside of the tire. Thus, to see if the constant circular motion had any effect on his tires David should measure the tread depth on both the inside and the outside of the tires before the experiment and measure the inside and the outside of the tires (at the same location on the tires) after the experiment. Then he can compare the tread loss on the inside of the tire to the tread loss on the outside of the tire.
Answers
w = width/distance
v = velocity
t = time
input the value:
where Bx = 3.3 X 10-6 T, By = 3.9 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.
1)
What is f, the frequency of this wave?
GHz
2)
What is I, the intensity of this wave?
W/m2
3)
What is Sz, the z-component of the Poynting vector at (x = 0, y = 0, z = 0) at t = 0?
W/m2
4)
What is Ex, the x-component of the electric field at (x = 0, y = 0, z = 0) at t = 0?
V/m
5)
Compare the sign and magnitude of Sz, the z-component of the Poynting vector at (x=y=z=t=0) of the wave described above to the sign and magnitude of SIIz, the z-component of the Poynting vector at (x=y=z=t=0) of another plane monochromatic electromagnetic wave propagating through vaccum described by:
B? =(BIIxi^?BIIyj^)cos(kz??t)
where BIIx = 3.9 X 10-6 T, BIIy = 3.3 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.
SIIz < 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)
SIIz < 0 and magnitude(SIIz) = magnitude(Sz)
SIIz > 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)
SIIz > 0 and magnitude(SIIz) = magnitude(
Answers
Final answer:
The question involves computation of frequency, intensity, Poynting vector and electric field of an electromagnetic wave, and comparison between two such waves. The solutions result in approximately: 10 GHz for frequency, 3.07 x 10^-12 W/m^2 for intensity, 1.3 X 10^-19 W/m^2 for the z-component of Poynting vector, and 1.43 V/m for the electric field. Moreover, the comparison yields that SIIz is less than zero and not equal to Sz in magnitude.
Explanation:
The subject of your question relates to
electromagnetic waves
and their properties such as frequency, intensity, Poynting vector, and the electric field component. These concepts belong to the realm of physics, and more specifically, are topics in the study of electromagnetic theory.
To solve your questions:
- The frequency f can be found using the formula: f = c/λ where c is the speed of light in vacuum (~3x10^8 m/s). For λ = 3 cm or 0.03 m, computation yields f ≈ 10^10 Hz or 10 GHz.
- The intensity I of an electromagnetic wave in a vacuum can be given by the equation I = 0.5*c*ε0*E^2, where E is the electric field amplitude. To compute I, first, we need to find E which is given by E = c*B, where B is the magnetic field amplitude. Here, B is the square root of (Bx^2 + By^2) resulting in approximately 4.77*10^-6 T. Thus, E ≈ 1.43 V/m and solving for I gives us I ≈ 3.07 x 10^-12 W/m^2.
- The z-component of the Poynting vector Sz at a specified point and time is given by Sz = E x H, where H = B/μ0, μ0 represents the permeability of free space. At t = 0, Sz = Ex*Hy - Ey*Hx = Ex*By, resulting in Sz ≈ 1.3 x 10^-19 W/m^2.
- The x-component of the electric field at t = 0 Ex≈1.43 V/m.
- Finally, comparing Sz of both waves (magnitudes and signs), we find that SIIz < 0 and the magnitude of SIIz does not equal the magnitude of Sz.
Learn more about Electromagnetic Waves here:
#SPJ12
Final answer:
The frequency of the wave is 10 GHz. While we can't expressly calculate the intesity, Sz, and Ex without more information, we can note that if the signs of Bx and By are swapped in a new wave, the Poynting vector would be flipped, hence SIIz would be negative and of equal magnitude to Sz.
Explanation:
An electromagnetic wave propagating through vacuum is described by certain electromagnetic fields which are associated with frequency, intensity, and Poynting vector which indicates the direction of energy flow. These can be calculated using certain formulas derived from wave equations.
Frequency can be acquired from the wavelength (λ) with the formula: f = c/λ, where c is the speed of light in vacuum. Using given λ = 3 cm, we get f = 10^10 Hz or 10 GHz.
The total Intensity (I) can be calculated as the average of the sum of the intensities in the x and y direction, given by: 1/2 ε_0 c E^2, where ε_0 is the permittivity of free-space and E is the electric field amplitude. However, more information might be needed to calculate this value. Similarly, without further information, we cannot calculate the exact values of Sz and Ex.
When comparing Sz and SIIz, if the signs of Bx and By are swapped in a new wave, this would flip the direction of the Poynting vector (since it is related to E × B), hence SIIz < 0 and its magnitude would still equal to Sz because the magnitudes of Bx and By do not change.
Learn more about Electromagnetic wave properties here:
#SPJ11
Answers
Answer:
The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s
Explanation:
Moment of inertia is given as;
I = ¹/₁₂×ML² + 2mr²
where;
I is the moment of inertia
M is the mass of the rod = 0.19 kg
L is the length of the rod = 0.43 m
m is the mass of the bead = 0.038 kg
r is the distance of one bead
Initial moment of inertial is given as;
Final moment of inertia is also given as
Angular momentum is the product of angular speed and moment of inertia;
= Iω
From the principle of conservation of angular momentum;
Given;
ωi = 12 rad/s
r₁ = 10.0 cm = 0.1 m
r₂ = 10.0cm/4 = 2.5 cm = 0.025 m
Substitute these values in the above equation, we will have;
Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s
Answers
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
Answers
Answer:
5.831 m/s
Explanation:
According to the work-energy law,
Work done between two points = Change in kinetic energy between the two points.
Since the plastic ball is initially at rest, its initial kinetic energy is 0 since the initial velocity = 0
Work done by the spring = ∫ F.dx
The spring is compressed by 10 cm, so, we integrate from -0.1 m to 0 m
Fₓ(x) = (-30.0 N/m)x+ (60.0 N/m²)x²
F = -30x + 60x²
W = ∫ F.dx = ∫ (-30x + 60x²) dx
W = [- 15x² + 20x³]⁰₋₀.₁ = 0 - [- 15(0.01) + 20(-0.001)] = 0.17 J
W = ΔKE
ΔKE = (mv²/2) - 0
mv²/2 = 0.17
m = 10 g = 0.01 kg
0.01 v² = 0.34
v² = 34
v = 5.831 m/s