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Nikitich [7]
4 years ago
6

A push broom is being pushed down across a rough floor. The broom moves to the right. What is the correct free body diagram of t

he broom?

Physics
2 answers:
zysi [14]4 years ago
5 0
The far right.

Fg is gravity which always acts down and since we assume the floor is flat the normal, Fn, acts opposite gravity, so straight up.

But you’re probably wondering about the pushing force, Fp, and the friction force, Ff. For the Fp, consider where the applied force is coming from. The head of the broom is on the floor and the man’s arms, where he’s applying the force from, is above and to the left, so when the man pushes the broom the force is down and to the right. The broom my not be moving down, but the applied force is still in that direction. And Ff always acts against motion so since the broom moves to the right, the friction is to the left.

Zielflug [23.3K]4 years ago
4 0

Answer:

D

Explanation: took the test on edg

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In an A.C circuit current leads voltage by phase π/2 then circuit is
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A car with a mass of 500 kg, gets off road and goes straight into a cliff of 30 m with an
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Answer:

The velocity of the car when it hits the ground is approximately 55.574 meters per second.

Explanation:

According to this expression, the car goes straight into the cliff and goes down due to gravity, whose situation is described by the Principle of Energy Conservation and supposing that all non-conservative forces are negligible:

U_{g,1}+K_{1} = U_{g,2}+K_{2} (1)

Where:

U_{g,1}, U_{g,2} - Gravitational potential energies of the car at the top and the bottom, measured in joules.

K_{1}, K_{2} - Translational kinetic energies of the car at the top and the bottom, measured in joules.

By definitions of gravitational potential and translational kinetic energies, we expand and simplify the equation above:

\frac{1}{2}\cdot m\cdot v_{2}^{2} = \frac{1}{2}\cdot m \cdot v_{1}^{2}+m\cdot g \cdot (z_{1}-z_{2}) (2)

v_{2}^{2} = v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})

v_{2} = \sqrt{v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})}

Where:

m - Mass, measured in kilograms.

v_{1}, v_{2} - Velocity of the car at the top and the bottom, measured in meters per second.

g - Gravitaitional acceleration, measured in meters per square second.

z_{1}, z_{2} - Height of the car at the top and at the bottom, measured in meters.

If we know that v_{1} = 50\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}}, z_{1} = 30\,m and z_{2} = 0\,m, then the velocity of the car when it hits the ground is:

v_{2} = \sqrt{\left(50\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (30\,m-0\,m)}

v_{2}\approx 55.574\,\frac{m}{s}

The velocity of the car when it hits the ground is approximately 55.574 meters per second.

7 0
3 years ago
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