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wlad13 [49]
4 years ago
12

A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he

has to catch up with the driver before he manages to get onto an expressway 60 km away?
Physics
1 answer:
Iteru [2.4K]4 years ago
6 0

Answer:

30.93 m/s

Explanation:

Given that, the speed of stolen car is,

v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, v_{p}

The policeman catches the stolen car at a distance of,

S=60km\\S=60000m

Now the distance covered by the policeman in time t is v_{p}\times t

And the distane cover by the thief in stolen car in time(t+60s) is v_{s}\times (t+60sec).

And these distances are equal and they are equal to 60000 m.

Therefore,

v_{p}\times t=v_{s}\times (t+60sec)=60000m

Therfore,

v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s

Now use this value to solve for minimum speed of policeman's car.

v_{p}\times 1940=60000\\v_{p}=30.93 m/s

Therefore minimum speed of policeman's car is 30.93 m/s.

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umka2103 [35]

Answer:

It may not be at the sea level

Explanation:

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In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo
Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find  the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Where, s = distance

t = time

a = acceleration

Put the value into the formula

60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2

a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}

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Hence, The acceleration of the both masses is 0.0244 m/s².

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As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave
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Answer:

-29.2\times 10^{3} N

Explanation:

We are given that

Mass of cars= m=1900 kg

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Final speed of car=v=0

Time=\Delta t=1.3 s

We have to find the average force exerted on the car.

Average force=\frac{change\;in\;momentum}{\Delta t}

F_{avg}=\frac{mv-mu}{1.3}

F_{avg}=\frac{1900(0)-1900(20)}{1.3}

F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N

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