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3 years ago

The source of the centripetal force that arises when a runner rounds a curve on a track is _____.

2 answers:

5 0

while taking a turn on a track, the track provides the frictional force acting towards the center of the curve. we know that to move in a curve, centripetal force is required. here since the frictional force acts towards the center, it provides the necessary centripetal force to move in circle. hence the correct choice is

friction

mel-nik [20]3 years ago

3 0

It's actually Friction.

I just did the test and got it right.

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A transport plane travelling at a steady speed of 132 ms and an altitude of 113 m, releases a parcel when it is directly above a

Crank

Answer:

what time you thinking. about coming down to take a break and ok I will get to?. that was a right answer?

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3 years ago

What is the length of the color attribute?

astraxan [27]

The length of the colour attribute is 6 characters.

<h3>What is a colour attribute?</h3>

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2 years ago

Which of the following is true of high clouds?

Wewaii [24]

I’m pretty sure it’s B

3 0

2 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

Read 2 more answers

The elastic energy stored in your tendons can contribute up to 35% of your energy needs when running. Sports scientists have stu

Anni [7]

Answer:

ΔE = 9.7mJ

Explanation:

See attachment below.

7 0

3 years ago