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liq [111]
3 years ago
5

WILL GIVE BRAINLIEST TO CORRECT ANSWER PLEASE HELP ME

Physics
1 answer:
koban [17]3 years ago
4 0

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

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While space walking, when the astronaut gets detached from the space ship, she floats in space holding a wrench. In order to get back to the spaceship, she should throw the wrench in the opposite direction of the spaceship. This action would  cause a reaction on her own body and she would be pushed away from the wrench and towards the spaceship. Thus, she can return back to the spaceship in this way.

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A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a
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Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

F=ma      Eqn. (1)

where

F : is the Net Force in Newtons ( N )

m : is the mass ( kg )

a  : is the acceleration ( m/s^2 )

We also know that the acceleration is denoted by the velocity ( v ) of an object as a function of time ( t ) with

a=\frac{v}{t}         Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

F=m\frac{v}{t}\\ \\Ft=mv     Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. Ft is in fact the Impulse  J of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ 22N and t=4 sec (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

J=Ft\\J=(22N)(4sec)\\J=88 kg m/s

So the impulse of the cart is J=88kg m/s

Then, we know that the cart is moving at v=10 m/s. Plugging in the values in Eqn. (3) we have:

(22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg

So the mass of the cart is m=8.8kg.

8 0
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