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3 years ago

11

Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?

1 answer:

Minchanka [31]3 years ago

6 0

For each load, Work = (mass) x (gravity) x (distance .

Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .

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Read 2 more answers

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

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4 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

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The frequency of oscillation is given by the formula as formula as follows :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg$

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3 years ago

A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop

KengaRu [80]

#A

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

$\\ \rm\longmapsto Force=ma$

$\\ \rm\longmapsto Force=4.4(-1.74)$

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#B

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

$\\ \rm\longmapsto a=\dfrac{v-u}{t}$

$\\ \rm\longmapsto u=v-at$

$\\ \rm\longmapsto u=0-(-1.74)(1.27)$

$\\ \rm\longmapsto u=1.74(1.27)$

$\\ \rm\longmapsto u=2.2m/s$

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3 years ago