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konstantin123 [22]
2 years ago
10

The following is regarding Momentum Change and Force. help is needed please!

Physics
1 answer:
mixer [17]2 years ago
4 0

The distance travelled  is 10 m and the velocity gained at the end of this time is 2 m/s.

<h3>Velocity of the object at the end of the time</h3>

F = mv/t

where;

  • m is mass of the object
  • v is velocity of the object
  • t is time

Ft = mv

v = Ft/m

v = (50 x 10)/250

v = 2 m/s

<h3>Distance traveled by the object</h3>

v² = u² + 2as

where;

u is initial velocity = 0

a is acceleration

a = F/m

a = 50 N/ 250 kg

a = 0.2 m/s²

v² = 0 + 2as

s = v²/2a

s = (2²)/(2 x 0.2)

s = 10 m

Thus, the distance travelled  is 10 m and the velocity gained at the end of this time is 2 m/s.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizont
kolbaska11 [484]

Answer:

The magnitude of the acceleration of the box is 2.01 m/s².

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following horizontal forces:

Fr = friction force.

Fx = Horizontal component of the applied force, F.

And we have the following vertical forces:

Fy = vertical component of the applied force.

N = normal force exerted on the box.

W = weight of the box.

According to Newton´s second law:

∑F = m · a

Then, in the horizontal direction:

Fx - Fr = m · a

Where "m" is the mass of the box and "a" its acceleration.

Fx can be obtained by trigonometry (see figure):

Fx = F · cos 30°

Fx = 90.0 N · cos 30°

Fr is calculated as follows:

Fr = μ · N

Where μ is the coefficient of friction and N the normal force.

So, we have to find the magnitude of the normal force.

Using Newton´s second law in the vertical direction:

∑F = N + Fy - W = m · a

Notice that the box has no vertical acceleration, then:

N + Fy - W = 0

Solving for N:

N = W - Fy

The weight is calculated as follows:

W = m · g

Where g is the acceleration due to gravity:

W = 20.0 kg · 9.8 m/s² = 196 N

And the vertical component of the applied force can be obtained by trigonometry:

Fy = F · sin 30°

Fy = 90.0 N · sin 30°

The normal force will be:

N = W - Fy = 196 N - 90.0 N · sin 30°

N = 151 N

Now, we can calculate the friction force:

Fr = μ · N

Fr = 0.250 · 151 N

Fr = 37.8 N

And now, we can obtain the acceleration of the box:

Fx - Fr = m · a

(Fx - Fr) / m = a

(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a

a = 2.01 m/s²

The magnitude of the acceleration of the box is 2.01 m/s².

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Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

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It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

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The volume of the system remains constant,

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Answer:

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