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pishuonlain [190]
3 years ago
11

A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to re

trieve the ball, you apply your brakes for 1.05 s. The car slows down from 15.0 m/s to 9.00 m/s. The mass of the car is 1250 kg. (a) During the time the brakes were applied, what was the average force (in N) exerted on your car?
Physics
1 answer:
ehidna [41]3 years ago
7 0

Answer:

F = 7.143 kN

Explanation:

given,

time taken to apply break = 1.05 s

car slows down from 15 m/s to 9 m/s

mass of the car = 1250 Kg

force is equal to the change in momentum with respect to time.

F = \dfrac{\Delta P}{t}

F = \dfrac{m(v_f - v_i)}{t}

F = \dfrac{1250\times (9 - 15)}{1.05}

F = \dfrac{-7500}{1.05}

F = -7142.85 N

F = - 7.143 kN

Force is acting opposite direction of velocity of car i.e. the sign is negative.

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A ball collides with a vertical, unmovable wall. There is no friction between the wall and the ball (the only force acting on th
user100 [1]

Answer: the same direction I.e to the left.

Explanation:

The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface will have the same speed (magnitude of velocity) as before the collision.

There will also be parallel force caused by friction, but it has to be treated separately for two reasons:

The perpendicular force is limited to coefficient of friction times the normal force. If that is not enough to stop the ball, it will skid on the surface.The perpendicular force, and this depends on the specific geometry, does not pass through the centre of mass of the ball. Therefore it imparts a moment on the ball that causes it to start rotating. And once the ball is rotating so that the point of contact is stationary, there is no momentum to cause any friction force anymore and the friction force disappears and stops decelerating the ball.

So what happens is that the vertical component of the velocity will be reversed, while the horizontal component will be somewhat reduced with the corresponding amount of kinetic energy transferred to energy of rotation. The rotation will always eliminate the friction force before the horizontal component of velocity is zeroed, so the ball will always continue in the same direction, just a bit slower.

If you instead threw an elastic box (which could not start rotating freely) it could actually bounce back.

7 0
3 years ago
Temperature of the resistor in an RC circuit. In this problem we will look at the temperature response of the resist
fomenos

RC circuit determines the capacitor's charging rate.

  • In RC (resistive and capacitive) circuits, a capacitor's time constant is the number of seconds required to charge it to 63.2% of the input voltage.
  • This duration is described by a single time constant. After two time constants, the capacitor will be charged to 86.5% of the input voltage.
  • The RC time constant, also referred to as tau, is the time constant (in seconds) of an RC circuit and is obtained by multiplying the circuit resistance (in ohms) by the circuit capacitance (in farads), This transient reaction time T is stated in terms of = R x C, where R is the resistor value in ohms and C is the capacitor value in farads.

Learn more about  RC circuits here brainly.com/question/13450553

#SPJ4.

4 0
2 years ago
A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h
babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

 f’= f₀  \frac{343}{343 + v_s}

Explanation:

This is a doppler effect exercise, where the sound source is moving

           f = fo \frac{v}{v-v)s}      when the source moves towards the observer

           f ’=f_o  \frac{v}{v+v_{sy}}  Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

          f’= f₀  \frac{343}{343 + v_s}

8 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
4 years ago
Anyone good with scientific notations? I sure am not
defon
B is the answer that I know of.
7 0
3 years ago
Read 2 more answers
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