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2 years ago

8

A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car, increasing the speed to 2 m/s. How much work did the

man do?
A. 1360 J
B. 360 J
C. 640 J
D. 1000 J

1 answer:

Vikki [24]2 years ago

5 0

Answer:

Option B - 360 J

Explanation:

One of the formulas used to find out how much work was done is :

E = 0.5×m×v² (where m is the mass and v is the velocity)

Now we substitute values from the question and simplify :

E = 0.5×500×1.2²

E = 250×1.2²

E = 250×1.44

E = 360 J

So our answer is Option B

Hope this helped and have a good day

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LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is:

$\bar F = \frac{m\cdot (v_{2}-v_{1})}{\Delta t}$

$\bar F = 12000\,N$

The average impact force is 12000 newtons.

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3 years ago

Will the 79 kg skier in the figure below slide down if f the coefficient of static friction is 0.25?

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Answer:

Man will not slide down

Explanation:

Given:

Coefficient of static friction = 0.25

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Computation:

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3 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

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Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em> (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

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