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3 years ago

A person on a bike (m=90kg) is traveling 4m/s at the top of a 2m hill. What is his/her kinetic energy?

Physics

2 answers:

andriy [413]3 years ago

7 0

Ke =(1/2 )mv^2
1/2(90*4)^2
720
Have a nice day

guajiro [1.7K]3 years ago

6 0

Answer:

Ek = 720J

Explanation:

Ek = 0.5mv^2

Ek = 0.5(90)(4)^2

Ek = 720J

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A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

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Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

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Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

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