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Alex777 [14]
3 years ago
8

Casey is playing with her yo-yo. She performs an "around the world" trick, in which the yo-yo travels up and around in a circle

before returning to her hand. At which point in its circular path would the yo-yo have the least potential energy?
Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

It has the least potential energy at the bottom of its circular path.

Explanation:

It has the least potential energy at the bottom of its circular path.

Remember the equation

U = m*g*h

where U is the potential energy

m is the mass of the yo-yo

h is the height

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Answer:

44.5s ; 22.64 m/s

Explanation:

The motion of the truck could be separated into 3 different phases :

First :

Time of motion :

Initial Velocity, u = 0 ; final velocity, v = 29 m/s

Acceleration, a = 2 m/s²

Recall: acceleration = change in velocity / time

Time = change in velocity / acceleration

Time = (29 - 0) / 2 = 14.5 second

Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m

Second :

Time = 25 seconds at constant speed

29 m/s for 25 seconds

v*t = 29 * 25 = 725 m

Third:

5 seconds before coming to rest

((29 + 0) /2) * 5

14.5 * 5 = 72.5 m

A.)

Length of journey = (14.5+ 25 + 5) = 44.5 seconds

B.)

Average velocity = total distance / total time taken

Average velocity = (210.25 + 725 + 72.5) / 44.5

= 1007.75 / 44.5

= 22.646067

= 22.64 m/s

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3 years ago
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When we move from 100x total magnification to 400x total magnification on a compound light microscope, what happens with resolut
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1. a) In order to sharpen the objective with a 40x objective (400x total magnification), the objective must be closer to the sample than with a 10x objective (100x total magnification), so the sample is out of focus.

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5 0
2 years ago
A bucket weighing 5 lbs is lifted at a constant rate from the bottom of a 100 ft well by a rope which weighs 5 lbs. The bucket h
8_murik_8 [283]

Answer:

W_{bucket} = 24934.85\,lbf\cdot ft

Explanation:

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W_{bucket} = (5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-0\,ft) + (25\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft) - (30\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (0\,ft)+(5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-50\,ft)

W_{bucket} = 24934.85\,lbf\cdot ft

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4 years ago
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