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bearhunter [10]
3 years ago
15

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

text for 4.0\,\text m4. 0m4, point, 0, start text, m, end text. How much kinetic energy does the crate gain?
Physics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

k_f = 40J

Explanation:

The work and energy theorem says that:

W_f =k_f-k_i

where W_f is the work of the force, k_f the final kinetic energy and k_i the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

Fd = k_f

where F is the force and d the distance. Then, replacing values, we get:

(10N)(4m) = k_f

40J = k_f

it means that the system gain 40J of kinetic energy.

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The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between
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Answer:

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Explanation:

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Now,

Q3 = 2 Q

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Divide equation (2) by equation (1), we get

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The mass of the car?
Nonamiya [84]

Answer:

1050 kg

Explanation:

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KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.

We can now substitute the values we know into this equation.

KE = 472 500 J and v = 30 m/s:

472 500 = 1/2 × m × 30²

Next, we can rearrange the equation to make m the subject and solve for m:

m = 472 500 ÷ (1/2 × 30²)

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m = 1050 kg

Hope this helps!

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