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Zielflug [23.3K]
3 years ago
8

A woman pushes a car with a force of 400 N for a distance of 15m. How much work has she done?

Physics
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

the work done by the woman is 6,000 J

Explanation:

Given;

force applied by the woman, F = 400 N

distance moved by the woman, d = 15 m

The work done by the woman is calculated as follows;

W = F x d

W = 400 N x 15 m

W = 6000 Nm = 6,000 J

Therefore, the work done by the woman is 6,000 J

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EX 6-1 A ball is twirled on a 0.870 - m-long string with a constant speed of 3.36 m / s . Calculate the acceleration of the ball
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Answer:

a=12.97\ m/s^2

Explanation:

Given that,

The length of a string, l = 0.87 m

Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2

So, the acceleration of the ball is 12.97\ m/s^2.

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3 years ago
During nuclear fission and fusion, matter that seems to disappear is actually converted into
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Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

Explanation:

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3 years ago
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
2 years ago
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