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ycow [4]
1 year ago
14

Velocity (m/s)

Physics
1 answer:
alekssr [168]1 year ago
3 0

Explanation:

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(a) <u>The</u><u> </u><u>segment</u><u> </u>A shows acceleration as velocity increases with the increase in time.

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(b) <u>The</u><u> </u><u>segment</u><u> </u>C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

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⠀

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(c) The velocity is segment B is <u>4</u><u>0</u><u>m</u><u>/</u><u>s</u><u>.</u> And in the diagram there is no change in velocity.

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⠀

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(d) The acceleration of segment B is <u>zero</u><u>.</u> As there in no change in curve and it is moving with uniform velocity.

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\:

<h2>Thank you!</h2>
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A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
2 years ago
A person walks the path shown below. The total trip consists of four straight-line paths.
dmitriy555 [2]

At the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

<h3>What is resultant displacement?</h3>

The resultant displacement of an object is the change in position of the object. It can be described as the shortest distance connecting the final position of the object to the initial position of the object.

<h3>Net horizontal displacement </h3>

Path 1 = 40 m

Path 2 = 0 m

Path 3 = 110 m x cos(30) = 95.26 m

Path 4 = 180 m x cos(60) = 90 m

Total horizontal displacement, X = 40 m + 0 m + 95.26 m + 90 m = 225.26 m

<h3>Net vertical displacement </h3>

Path 1 = 0 m

Path 2 = 230 m

Path 3 = 110 m x sin(30) = 55 m

Path 4 = 180 m x sin(60) = 155.885 m

Total horizontal displacement, Y = 0 m + 230 m + 55 m + 155.885 m = 440.885 m

<h3>Resultant displacement</h3>

R = √(X² + Y²)

R = √(225.26² + 440.885²)

R = 495.1 m

<h3>Direction of the displacement</h3>

θ = arc tan (Y/X)

θ = arc tan (440.885/225.26)

θ =  63⁰

Thus, at the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

Learn more about resultant displacement here: brainly.com/question/13309193

#SPJ1

3 0
10 months ago
Which of the following is NOT usually published with a scientific report?
dusya [7]

C cost (I need points)

5 0
3 years ago
A car starts from rest and drives at 45 m/s in 20 s. how far did it travel
Schach [20]

Answer:

distance=speed×time.

45×20=900m.

6 0
3 years ago
You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building.
DanielleElmas [232]

Answer:

   t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

         v = v₀ + a t₁

the elevator starts with zero speed

         v = a t₁

         t₁ = v / a

         t₁ = 18.8 / 2.40

         t₁ = 7.833 s

in this time he runs

         y₁ = v₀ t₁ + ½ a t₁²

         y₁ = ½ a t₁²

         y₁ = ½ 2.40 7.833²

         y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

           x_total = x₁ + x₂

           x₂ = x_total - x₁

           x₂ = 373 - 73,627

           x₂ = 299.373 m

this distance travels at constant speed,

           v = x₂ / t₂

           t₂ = x₂ / v

           t₂ = 299.373 / 18.8

           t₂ = 15.92 s

therefore the total travel time is

           t_total = t₁ + t₂

           t_total = 7.833 + 15.92

           t_total = 23.757 s

6 0
3 years ago
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