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dlinn [17]
4 years ago
10

I need help on a physics problem!! Please help

Physics
1 answer:
Ksivusya [100]4 years ago
5 0
What you need help with?
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which of the following are vector quantities? check all that apply. a. acceleration b. mass c. displacement d. force
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All are vector quantities except b. mass
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A block of gelatin is 120mm by 120mm by 40mm when unstressed. A force of 49N is applies tangentically to the upper surface causi
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120x JB y and the x got to the s
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A physical therapy exercise has a person shaking a 5.00 kg weight up and down rapidly. if the barbell is moving at 4.50 m/s, wha
Ivanshal [37]

The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

<h3>Magnitude of required force to stop the weight</h3>

The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;

F = ma

F = m(v/t)

F = (mv)/t

F = (5 x 4.5)/0.333

F = 67.6 N

Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

Learn more about force here: brainly.com/question/12970081

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7 0
2 years ago
Make the following prefix conversions.<br> 0.001s =ms
Free_Kalibri [48]

Answer:

To convert a millisecond measurement to a second measurement, divide the time by the conversion ratio. The time in seconds is equal to the milliseconds divided by 1,000.

Explanation:

hope it helps

4 0
2 years ago
A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab
Darya [45]

Answer:

a. F=126N

b. E_K=1323J

Explanation:

Given:

m=54kg

v=7 m/s

t= 3s

The runner force average to find given the equations

a.

F=m*a

a=\frac{v}{t}

F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}

F=126N

b.

Work done by the system by this force so

W=F*d

W=E_K

E_K=\frac{1}{2}*m*v^2

E_K=\frac{1}{2}*54kg*(7m/s)^2

E_K=1323J

6 0
3 years ago
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