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2 years ago

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Assume that a person bouncing a ball represents a closed system. Which statement best describes how the amounts of the ball's po

tential and kinetic energy change?
A. Both the kinetic and potential energy of the ball increase
B.The ball's potential energy decreases as it rises and it's kinetic energy increases
C. The ball's kinetic energy increases as it falls and its potential energy decreases
D. Both the kinetic and potential energy of the ball decrease

PLEASE HELP ME ASAP THIS IS IMPORTANT

1 answer:

Zinaida [17]2 years ago

7 0

Answer:

C.

Explanation:

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A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

enyata [817]

Answer:

Emitted power will be equal to $7.85\times 10^{-5}watt$

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing $I=10^{-12}W/m^2$

We have to find the emitted power

Emitted power is equal to $P=I\times A$

$P=I\times 4\pi R^2$

$P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt$

So emitted power will be equal to $7.85\times 10^{-5}watt$

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2 years ago

"As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situa

____ [38]

Answer:

$\frac{I_{2}}{I_{1}} = 0.04$

Explanation:

The intensity of a star noticed at a certain distance is inversely proportional to the square of distance. Then:

$I_{1}\cdot r_{1}^{2} = I_{2}\cdot r_{2}^{2}$

The intensity of the Sun in Jupiter relative to Earth is:

$\frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}$

$\frac{I_{2}}{I_{1}} = \left(\frac{1\,AU}{5.2\,AU} \right) ^{2}$

$\frac{I_{2}}{I_{1}} = 0.04$

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2 years ago

A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive dir

givi [52]

Answer:

$\frac{1}{8} y'' + 2y' + 24y=0$

Explanation:

The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by

$my'' + \zeta y' + ky=0$

Where m is the mass, ζ is the damping constant, and k is the spring constant.

The spring constant k can be found by

$w - kL=0$

$mg - kL=0$

$4 - k\frac{1}{6}=0$

$k = 4\times 6 =24$

The damping constant can be found by

$F = -\zeta y'$

$6 = 3\zeta$

$\zeta = \frac{6}{3} = 2$

Finally, the mass m can be found by

$w = 4$

$mg=4$

$m = \frac{4}{g}$

Where g is approximately 32 ft/s²

$m = \frac{4}{32} = \frac{1}{8}$

Therefore, the required differential equation is

$my'' + \zeta y' + ky=0$

$\frac{1}{8} y'' + 2y' + 24y=0$

The initial position is

$y(0) = \frac{1}{2}$

The initial velocity is

$y'(0) = 0$

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2 years ago

A rocket train car that is 30 m long is traveling from Los Angeles to New York at v=0.5c when a light at the center of the car f

Nataly [62]

Answer: The reference frame of a passenger in a seat near the center of the train

Explanation:

the speed of light is the same for the passenger and the bicyclist

then the avents are simultaneous fo the passenger not for the bicyclist

the delay between the two events for the bicyclist is

Δt=Δd/vs

where

Δd= lenght of train

vs=speed of sound

the reference frame of a passenger in a seat near the center of the train

Solution:

The space and time transformations are:

x' = γ(x - vt)

t' = γ(t - vx/c²).

In the primed frame the two events are simultaneous, so that Δt' = 0. Also here Δx' = 30. In the unprimed frame Δx' = 30 = γ(Δx - v Δt).......(*)

We also have Δt' = 0 = γ(Δt - vΔx/c²)→Δx = c²Δt/v......(**)

Substituting (**) in (*): 30 = γ(c²Δt/v - vΔt))→Δt = 30/(c²/v - v) =

30/(2c - 0.5c) = 6.7 x 10^(-8)s

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2 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

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5 0

2 years ago