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3 years ago

A 1.00x10^3 crate is being pushed across a level floor at a constant speed by a force

Physics

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Answer:

The answer is 0.39 m/s ²

Explanation:

Since the box moves with steady speed,

Friction=force in even direction=260*Cos20=244.3N

Ordinary force=mg+FSinθ=1089N

μN=f

μ=244.3/1089=0.224

N=mg-FSinθ=911.1N

a=(260 Cos20-0.224*911.1)/(1000/9.8)=0.39m/s²

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A ski jumper has a mass of 59.6 kg. She is moving with a speed of 23.4 m/s at a height of 44.6 meters above the ground. Determin

Daniel [21]

Hello!

Use the formula:

M = k * p

Data:

M = Mechanic energy

k = Kinetic energy

p = Potencial energy

Descomposing:

M = (0,5*mv²) + (mgh)

Replacing:

M = (0,5 * 59,6 kg * (23,4 m/s)²) + (59,6 kg * 9,81 m/s² * 44,6 m)

M = 16317,28 J + 26076,54 J

M = 42393,82 J

The mechanic energy is <u>42393,82 Joules.</u>

7 0

3 years ago

A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

Dmitriy789 [7]

Answer:

$A = 0.081 \ m$

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

The mass is $m = 0.750 \ kg$

The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=> $A = 0.081 \ m$

Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

3 0

3 years ago

What could result from regular participation in sports and exercise

shusha [124]

Common health issues that can be positively affected, prevented or controlled by exercise.

5 0

3 years ago

At what point will the electric field of a charged object be strongest?

puteri [66]

Answer:

The answer is C because they have to be close to be able to interact

7 0

3 years ago

A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

$v_x=149\ m/s$

In the y axis

$v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}$

The resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s$

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

4 0

4 years ago