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3 years ago

15

Which of the following represents a case in which you are not accelerating?a. Driving in a straight line at 60 miles per hr.b. G

oing from 0-60 mph in 10 secc. Slamming on the brakes to come to a stop at a stop signd. Driving 60 miles per hour around a curve

1 answer:

Setler79 [48]3 years ago

6 0

C. Slamming on the brakes to come to a stop at a stop sign. Here you are Decelerating.

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Alguien que sepa de electromecánica porfavor

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2 years ago

a light-year is ____ A. the amount of time it takes light to get to the nearest star B. the distance light travels in a day C. t

Vinil7 [7]

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3 years ago

Read 2 more answers

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

I need HELP Please !!

aliya0001 [1]

You have to figure it out

5 0

2 years ago

How do you find the capacitance in this?

Lostsunrise [7]

Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

Ceq = 11C/6

3 0

2 years ago