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leva [86]
3 years ago
14

To test the resiliency of its bumper during low-speed collisions, a 2 840-kg automobile is driven into a brick wall. The car's b

umper behaves like a spring with a force constant 6.00 106 N/m and compresses 3.98 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall
Physics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

V= 1.82 m/s

Explanation:

Given that

mass , m = 2840 kg

Spring constant ,K = 6 x 10⁶ N/m

Compression in the spring , x = 3.98 cm

Lets take speed of the car before impact = V

Now by using energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mV^2

Kx^2=mv^2

V=\sqrt{\dfrac{Kx^2}{m}}

Now by putting the values in the above equation

V=\sqrt{\dfrac{6\times 10^6\times (3.98\times 10^{-2})^2}{2840}}

V= 1.82 m/s

Therefore the speed of the car before impact will be 1.82 m/s.

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
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Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power  ,

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final speed of the car, v=32 mph

At full power,

initial speed of the car, v_0=0

final speed of the car, v=64 mph

a)

At constant power, KE = \frac{1}{2} mv^2

At full power, KE = \frac{1}{2} m(2v)^2

So KE_f = 4KE_i

So, time to reach 64 mph speed is 4 times more than the initial time

t = 4*1.10 =4.40 s

b)

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For final 64 mph speed,

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