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Tatiana [17]
3 years ago
12

An automobile traveling 50km/hr. It decelerates at 5 meter per second square. How long will it take to stop the car? How far wil

l the car travel after a brakes applied?
Physics
1 answer:
const2013 [10]3 years ago
3 0
We know, a = v/t
Here, a = 5 m/s²
v = 50 km/h= 13.88 m/s

Substitute their values into the expression:
5 = 13.88 / t
t = 13.88/5
t = 2.78 sec

Now, we know, v = d/t
13.88 = d/2.78
d = 13.88 * 2.78
d = 38.53 meter

In short, Your Answers would be:
i) It will take 2.78 sec
ii) It will travel for 38.53 m after a brakes applied.

Hope this helps!
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A 2 200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.20 m before coming into contact
77julia77 [94]

Answer:

F = 6.10 \times 10^5 N

Explanation:

Here we know that pile will go down under gravity for d = 4.20 m before coming in contact with beam

Now after being in contact with the beam the pile further go down by L = 15.4 cm.

Finally it will come to rest.

So here we can say that work done by gravity + work done by beam on the pile must be equal to change in its kinetic energy

So here we have

mg(L + d) - F(L) = 0

F = \frac{mg(L + d)}{L}

F = \frac{2200\times 9.81(4.20 + 0.154)}{0.154}

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8 0
3 years ago
We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (
kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 I =  476 \ N \cdot s

b

 I_1 =  14.21 \  N\cdot s

c

    F  = 20300 \  N

Explanation:

Considering the first question

From the question we are told that

   The force produced is F  =  3400 \ N

   The duration of the punch is  t =  0.14 \  s

Generally the impulse delivered is mathematically represented as

    I =  F  *  t

=>    I =  3400  *  0.14

=>    I =  476 \ N \cdot s

Considering the second  question

   The approaching velocity of the ball is  v_b  =  45 \ m/s

    The leaving  velocity of the ball is  v_l  =  -53 \ m/s

     The mass of the ball is  m_b  =  0.145 \  kg

Generally the magnitude of the impulse delivered is mathematically represented as

     I_1 =  m*  v_b  - m *  v_l

=>     I_1 =  [0.145 *  45]  - [0.145 * -53]

=>     I_1 =  14.21 \  N\cdot s

Considering the third  question

     The  duration of the impact of the bat is  t _1 =  0.7 \ ms  =  0.7 *10^{-3} \  s

      Generally the average force exerted by the bat is mathematically represented as  

       F  =  \frac{I_1}{t_1}

=>     F  =  \frac{14.21 }{0.7 *10^{-3}}

=>       F  = 20300 \  N

 

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