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3 years ago

A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:

1 answer:

8 0

To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.

PART A ) The Impulse can be calculcated as follows

$L= F\Delta t$

Where,

F = Force

$\Delta t =$ Change in time

Replacing,

$L = (35N)(0.2s)$

$L= 7N\cdot s$

PART B) At the same time the momentum follows the conservation of momentum where:

Initial momentum= Final momentum

And the change in momentum is equal to the Impulse, then

$\Delta p = L$

And

$\Delta p = p_f - p_i$

There is not initial momentum then

$\Delta p = p_f$

$L = p_f$

$p_f = 7N\cdot s = 7kg\cdot m/s$

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If the 50-kg crate starts from rest and achieves a velocity of v = 4 m/s

Kay [80]

Answer:

P = 227 N

Explanation:

Assuming the crate is on horizontal ground and subject to a horizontal force.

F = ma

P - μmg = ma

P = m(a + μg)

P = m(v²/2s + μg)

P = 50(4²/(2(5))+ 0.3(9.8))

P = 227 N

6 0

2 years ago

If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit

Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) = (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 = (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0

3 years ago

A ball is thrown upward from the ground with an initial speed of 22.0 m/s; at the same instant, another ball is dropped from a b

Vikki [24]

They will be together in 8 minutes

3 0

3 years ago

A small mirror is attached to a vertical wall, and it hangs a distance of 1.87 m above the floor. The mirror is facing due east,

Oksana_A [137]

Answer:

t = 1.62 h

Explanation:

A flat mirror fulfills the law of reflection where the incident angle is equal to the reflected angle.

θ_i = θ_r

If we use trigonometry to find the angles, the mirror is at a height of L = 1.87 m, and the reflected rays reach a distance x1 = 3.56 m

tan θ₁ = x₁ / L

tan θ₁ = $\frac{3.56}{1.87}$

θ₁ = tan⁻¹ 1.90

θ₁ = 62.29º

for the second case x₂ = 1.46 m

tan θ₂ = x₂ / L

θ₂ = tan⁻¹ $\frac{1.46}{1.87}$

θ₂ = 37.98º

the difference in degree traveled is

Δθ = θ₁- θ₂

Δθ = 62.29 - 37.98

Δθ = 24.31º

as in the exercise they indicate that every 15º there is an hour

t = 24.31º (1h / 15º)

t = 1.62 h

4 0

3 years ago

How does kinetic energy affect the stopping distance of small vehicle compared to a large vehicle

Nadusha1986 [10]

The kinetic energy for a large vehicle is different from that of a smaller vehicle, assuming that the vehicles are travelling at the same speed and stopping the same distance. This is because for a larger vehicle the kinetic energy is higher, as the mass for a larger vehicle, is more than the smaller vehicle.

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3 years ago