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kati45 [8]
3 years ago
6

Agent Bond is standing on a bridge, 17 m above the road below, and his pursuers are getting too close for comfort. He spots a fl

atbed truck approaching at 28 m/s , which he measures by knowing that the telephone poles the truck is passing are 22 m apart in this country. The bed of the truck is 1.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he jumps down from the bridge onto the truck, making his getaway.
Physics
1 answer:
faust18 [17]3 years ago
6 0

Answer:

2 poles

Explanation:

We are told Agent Bond is standing on a bridge, 17 m above the road below.

Now, the bed of the truck is 1.5 m above the road.

It means the distance hewill need to fall = 17 - 1.5 = 15.5 m

acceleration due to gravity = 9.8 m/s²

Initial velocity = 0 m/s

Thus, to find the time, we will use the equation;

s = ut + ½at²

Plugging in the relevant values gives;

15.5 = 0 + ½9.8t²

Multiply through by 2 to give;

15.5 × 2 = 9.8t²

31 = 9.8t²

t² = 31/9.8

t = √(31/9.8)

t = 1.78 sec

We are told that he spots a flatbed truck approaching at 28 m/s and that the telephone poles are 22m apart.

Thus; The truck will pass the poles at a rate of; Velocity/distance between poles = 28/22 = 1.273 poles per second

Since, we got the time to be 1.78 seconds, then we can find the number of poles by multiplying it with the rate of poles per seconds.

Thus;

Number of poles = 1.78 second × 1.273 poles/ seconds = 2.27 poles

This is approximately 2 poles

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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

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S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
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30/ 3.2 = 9.38
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