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andrey2020 [161]
3 years ago
11

A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2

r. The magnitude of the gravitational force exerted by the planet on Moon 2 is which of the following compared with the gravitational force exerted by the planet on Moon 1?
half as largetwice as largeone-fourth as largefour times as largethe same
Physics
1 answer:
saw5 [17]3 years ago
7 0

Answer:

Half as large.

Explanation:

Using Newton's law of universal gravitation, if the mass of the planet is <em>M</em> and of the Moons 1 and 2 is <em>m</em>, them the force exerted by the planet on them will be:

F_1=\frac{GMm}{r}

F_2=\frac{GMm}{2r}

Which clearly shows that the force that the planet exerts on the Moon 2 is half  the force it exerts on the Moon 1.

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Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration
LenaWriter [7]

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration,a=\frac{v-u}{t}

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a=\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2

Acceleration=a=5.6 m/s^2

Hence, the acceleration of cheetah=5.6m/s^2

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Which of the following is evidence that supports the idea of uniformitarianism?
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D. rates of soil erosion are much lower during droughts that last several years
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No sistema representado abaixo, a massa do bloco A é igual a 2 kg, a do bloco B é igual a 8 kg.
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A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
3 years ago
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