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3 years ago

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a car mass 1500kg traveling at a uniform velocity of 30 m/s due east .the driver applies the brake to slow down the car to a vel

ocity of 10 m/s east . what is the work done in slowing down the car ?

1 answer:

bonufazy [111]3 years ago

3 0

Answer:

this is the answer

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A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

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Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

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3 years ago

A dragster completed a 402.3-(0.2500-mi) run in 5.023s. If the car had a constant acceleration, what was its acceleration and fi

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Answer:

v = 80.092 m/s

a = 15.945 m/s²

Explanation:

Given,

The distance completed by the dragster, d = 402.3 m

The time taken by the dragster to complete that distance is, t = 5.023 s

The initial velocity of the dragster, u = 0

The final velocity of the dragster, v = ?

The acceleration of the dragster, a = ?

The velocity of the body is given by the formula

v = d/t m/s

Substituting the above values inn the equation

v = 402.3 m / 5.023 s

= 80.092 m/s

So, the velocity of the dragster is, v = 80.092 m/s

The acceleration of the body is given by the formula

a = (v - u)/t m/s²

Substituting the above values in the equation

= (80.092 - 0) / 5.023

= 15.945 m/s²

Hence, the acceleration of the dragster is, a = 15.945 m/s²

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Distance and angular size

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