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3 months ago

how fast would a(n) 73 kgkg man need to run in order to have the same kinetic energy as an 8.0 gg bullet fired at 430 m/sm/s ?

Physics

1 answer:

Mars2501 [29]3 months ago

8 0

Answer:

an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,

v=4.0m/s

Explanation:

Typically, the mathematical formula for kinetic energy is as follows:

KE= $\frac{1}{2} *m*v^{2}$

K.E= kinetic energy

M=mass

V= speed

so,

KE=1/2*0.008*(420)*420

K.E=672.4J

Therefore,

v*v=672.4/85

v*v=15.821

v=4.0m/s

Therefore

v=4.0m/s

What is Kinetic Energy?

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

To know more about Kinetic Energy? Visit to: -

brainly.com/question/22174271

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The y-position of a damped oscillator as a function of time is shown in the figure.

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(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) = e^(-bx)

ln[-0.857 / cos(ω)] = -bx

ln[-0.857 / cos(ω)] / b = - x ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx

ln[-0.57 / cos(2ω)] /2b = - x ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) = cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω - cosω - 6 = 0

let cosω = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω = arc cos(-0.92)

ω = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

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