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3 years ago

A Bicyclist that has a mass of 75 kg and is traveling at a velocity of 15 m/s would have how much kinetic energy ?

Physics

1 answer:

Marrrta [24]3 years ago

5 0

Answer:

8437.5

Explanation:

By applying the formula of kinetic energy:

Kinetic energy=1/2 mv^2

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Two everyday examples of a starting motion

Tasya [4]

Pushing, pulling is the answer

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3 years ago

I need help it is about physics

VMariaS [17]

Answer:

a force

Explanation:

Because if we apply force then only an object can slow down, speed up or change direction

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3 years ago

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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

3 0

3 years ago

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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

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3 years ago

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How can the frequency with which the direction of a current changes be regulated?

sesenic [268]

Answer:

In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.

Explanation:

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3 years ago