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3 years ago

If you are in a rear-wheel skid, you must release your brakes and turn your steering wheel __________

1 answer:

3 0

I think the correct answer would be A. If you are in a rear-wheel skid, you must release your brakes and turn your steering wheel into the direction of the skid. A skid in vehicles is a situation wherein one or more wheels are slipping in relation to the road and affecting the whole handling of the vehicle. To do a rear wheel skid, you should be releasing the break with caution and move or steer the wheel into the direction you would want to move.

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Which of the following best defines spring constant ?

katovenus [111]

Answer:

Explanation:

7 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

2 years ago

Read 2 more answers

What force is needed to propel a 421-kg car with an acceleration of 3.77 m/s2?

nekit [7.7K]

brainly.com/question/11542618?answering=true&answeringSource=greatJob%2FquestionPage

7 0

2 years ago

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.75 m/s. If the roof is pitched at

dalvyx [7]

Answer:

a) t = 0.528 s

b) D = 1.62 m

Explanation:

given,

speed of the baseball = 3.75 m/s

angle made with the horizontal = 35°

height of the roof edge = 2.5 m

using equation of motion

$s = ut +\dfrac{1}{2}gt^2$

$2.5 = vsin\theta \ t +\dfrac{1}{2}gt^2$

$2.5 = 3.75\ sin35^0 \ t +\dfrac{1}{2}\times 9.8 t^2$

4.9 t² + 2.15 t - 2.5 = 0

on solving the above equation

t = 0.528 s

b) D = v cos θ × t

D = 3.75 × cos 35° ×0.528

D = 1.62 m

3 0

3 years ago

If two objects at different temperatures are in contact with each other, what happens to their temperatures?

miv72 [106K]

In that case, heat energy flows from the warmer object to the cooler one.
As heat flows from one to the other, the temperature of the warmer object
falls, and the temperature of the cooler object rises. When the temperatures
are equal, the flow of heat energy from one to the other stops.

6 0

2 years ago