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melomori [17]
3 years ago
5

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

, in astronomical units, AU. If planet Y is twice the mean distance from the sun as planet X, by what factor is the orbital period increased? A. 2^1/3. B. 2^1/2. C. 2^2/3. D. 2^3/2
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

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damaskus [11]

Answer:

0.8J

Explanation:

Given parameters:

Force  = 20N

Compression  = 0.08m

Unknown:

Spring constant  = ?

Elastic potential energy  = ?

Solution:

To solve this problem, we use the expression below:

           F = k e

F is the force

k is the spring constant

e is the compression

             20  = k x 0.08

              k  = 250N/m

Elastic potential energy;

       EPE  = \frac{1}{2} k e²    =  \frac{1}{2}  x 250 x 0.08²

 Elastic potential energy = 0.8J

3 0
3 years ago
A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I
vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

\epsilon = \frac{d \phi}{dt}

\int E.dl = \frac{d(BA)}{dt}

E(2\pi r)= \pi r^2 \frac{dB}{dt}

E = \frac{r}{2} \frac{dB}{dt}

electric field at point P_1 as follow

E = \frac{r}{2} \frac{dB}{dt}

E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}

E = 6.3\times 10^{-6} V/m

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

a = \frac{qE}{m}

a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}

a = 603.59 m/s^2

5 0
3 years ago
How do you know if you have all the forces needed for a FBD?
Korolek [52]
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If a cart of a roller coaster has a mass of 250kg and is at a height of 14 meters. What is the cart's potential energy?
ahrayia [7]

Answer:

3430000 J

Explanation:

The formula for potential energy is PE=mgh.

M being the mass, g being the force of gravity, and h being the height.

First thing you want to do is convert 250 kg to g (grams).

From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).

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3 years ago
What is the average acceleration of a car that starts from
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Answer:

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Explanation:

In the picture above.

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