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BabaBlast [244]
3 years ago
14

A 64.0 kg daredevil exhibitionist hangs on to the end of a long bungee cord attached to the railing of the Brooklyn Bridge and o

scillates vertically. (Do NOT try this!) The spring constant of the bungee cord is 113 N/m, and the system's damping constant is 0.113 kg/s.
How long does it take for the oscillation amplitude to decrease to 93.8% of its initial value?
What is the frequency of oscillation?
Physics
1 answer:
Mashutka [201]3 years ago
8 0

Answer: t=72.5secs and f= 0.194Hz

Explanation:

0.938=e^[-bt/(2M)]

ln(0.938) = -(0.113 kg/s)(t) / [2(64.0 kg)]

t = 72.5 s

ω = sqrt[k/m - {b/(2M)}²]

ω = sqrt[(113 N/m)/(64.0 kg) - {(0.113kg/s)/(2(64.0 kg))}²]

ω = 1.33 rad/s

2πf = 1.33 rad/s

f = 0.194 Hz

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3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
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a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

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3 years ago
A yellow and green car traveled 400 miles to Dayton, OH. The green car made the trip in 10 hours. The yellow arrived in 8 hours.
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Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

Explanation:

The speed of each car is defined as:

v=\frac{d}{t}

where d is the distance traveled by the car and t is the time taken.

For the yellow car, d=400 mi and t=8 h, so its speed is

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For the green car, d=400 mi and t=10 h, so its speed is

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So, the correct choice is

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

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Required: Kinetic energy K.E = ?

Formula: K.E = 1/2 mv²

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Answer:

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