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Ratling [72]
3 years ago
6

Sarah weighs 392 N. She pushes a 56 kg rock with a force of 65 N. What force does the rock exert on Sarah?

Physics
1 answer:
Leona [35]3 years ago
6 0

Answer:

The force the rock exerts on Sarah =#is 65 N

Explanation:

The given parameters are;

Sarah's weight = 392 N

The force with which Sarah pushes the rock = 65 N

The mass of the rock = 56 kg

The weight of the rock = The mass of the rock × Acceleration due to gravity

∴ The weight of the rock = 56 kg ×9.81 m/s² = 549.36 N

Given that the force Sarah applies to push the rock = 65 N, then by Newton's third law of motion which states that action and reaction are equal and opposite, the force that the rock exert on Sarah is equal an opposite to the force Sarah is applying

Therefore, the force the rock exerts on Sarah = 65 N (in the opposite direction).

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A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
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Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
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