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Brrunno [24]
3 years ago
7

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

elocity of 2rev/s. During the dismount she stretches out into the straight position, with a length of 1.5m, (assume she is a uniform rod through the center) for her landing. The gymnast has a mass of 50kg. What is her angular velocity in the straight position?
Physics
1 answer:
kifflom [539]3 years ago
6 0

Here we will say that there is no external torque on the system so we will have

L_i = L_f

here we know that

L_i = I_1\omega_1

where we know that

I_1 = \frac{2}{5}mr^2

Also we know that

I_2 = \frac{1}{12}mL^2

initial angular speed will be

\omega_1 = 2\pi(2rev/s) = 4\pi rad/s

now from above equation

\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega

0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega

0.452 = 0.1875 \omega

now we have

\omega = 2.41 rad/s

so final speed will be 2.41 rad/s

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Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an
Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0
3 years ago
The tired of a car support the weight of a stationary. If onetire has a slow leak, the air pressure within the tire will_____wit
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Answer:

d) Decrease, increase, remain constant

Explanation:

If one tire has a slow leak, the air pressure within the tire will_DECREASE____with time due to outflow of air , the surface area between the tire and the road will__INCREASE__in time,due to flattening of tire.

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Answer:

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Explanation:

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8 0
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