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kifflom [539]
3 years ago
13

The acceleration due to gravity on planet a is one-sixth what it is on planet b, and the radius of the planet a is one-fourth th

at of planet
b. the mass of planet a is what fraction of the mass of planet b?
Physics
1 answer:
nevsk [136]3 years ago
8 0
The acceleration due to gravity on a large body is given by:
acceleration = (gravitational constant * mass of body) / radius²

For b, this is:
g = GM/r²
M = gr²/G

For a,
g(a) = 1/6 g
r(a) = 1/4 r
M(a) = g(a)r(a)²/G
M(a) = 1/6 g * (1/4 r)²/G
M(a) = gr²/96G

The fraction is given as:
M(a) / M = (gr²/96G) / (gr²/G)
M(a) / M = 1/96

The mass of planet a is 1/96 of the mass of planet b.
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If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob
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Answer:

Explanation:

Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:

T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m

By using the y=A\sin(\omega t)  \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}

The mean position is the position when <em>y</em> = 0, so:

\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi rad/s

and v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi} in centimeters (cm).

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How much force is exerted if a 250 kg object has an acceleration of 750 m/s2 ?
Andrew [12]
Use Newton’s second law: F=ma
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6 0
4 years ago
Read 2 more answers
Find the magnitude &amp; direction: 50N 25 N 35 N 10N
adoni [48]

F1x + F2x = Rx

↓

Rx = F1x + F2x

↓

Rx = F1 cos45° + F2

↓

Rx = (50N)(cos45°) + 60N

↓

Rx = 95N

Similarly, if we sum all the y components, we will get the y component of the resultant force:

F1y + F2y = Ry

↓

Ry = F1y + F2y

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Ry = F1 sin45° + 0

↓

Ry = F1 sin45°

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Ry = (50N)(sin45°)

↓

Ry = 35N

At this point, we know the x and y components of R, which we can use to find the magnitude and direction of R:

Rx = 95N

Ry = 35N

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3 years ago
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