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jeka57 [31]
3 years ago
7

A weight lifter struggles but manages to keep a heavy barbell abovehis head. Occasionally he slips and the barbell starts to fal

l downward, but he always recovers. In each case, compare the force exerted by the weight lifter on the barbell to that exerted by the barbell on the weight lifter.
A)- With the barbell at rest.
B)- With the barbel moving upward.
C)- With the barbel moving downward.
Physics
2 answers:
shutvik [7]3 years ago
4 0
A) The lifter's and the barbell's force are equal and opposite in direction
B) The lifter's force is greater than that of the barbell and in the upward direction
C) The barbell's force is greater than that of the lifter and in the downward direction
lakkis [162]3 years ago
3 0

Explanation:

(A)

-With the barbell at rest- Both forces are equal

When the barbell is at rest then in this condition both forces are equal. The force exerted by the weight lifter on the barbell is equal to the force exerted by the barbell on the weight lifter.

(B)

-With the barbell moving upward- Forces are equal

When the barbell is moved upward by the weight lifter then the force exerted on the barbell by the weight lifter is greater than the gravitational force exerted on the barbell.

(C)

-With the barbell moving downward-Forces are equal

When the barbell is moved downward by the weight lifter then the force exerted on the barbell due to gravity is greater than the force exerted by the weight lifter on the barbell.

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M = W/g
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Which list of elements contains only metals?
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In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b
lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting  frictional force is provided by Centripetal force

f=\mu N=\mu m\omega ^2\cdot r

weight of object=mg

Equating two

f=mg

\mu m\omega ^2\cdot r=mg

\omega ^2=\frac{g}{\mu r}

\omega =\sqrt{\frac{g}{\mu r}}

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8 0
3 years ago
An elephant and a mouse would both have zero weight in gravity-free space. If they were moving toward you with the same speed, w
Dovator [93]

The elephant and the mouse having zero weight in a gravity free space will not bump into you at the same effect.

<u>Explanation: </u>

When both are in a gravity free space, the weights are zero, as we know that the\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}

\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero

\text {similarly,weight of mouse}=\text {mass of mouse } \times \text {zero gravity}=zero

But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.  

\text { momentum of elephant }=\text { mass of elephant } \times v  \text { momentum of mouse = mass of mouse } \times v

And as we know \text { mass of elephant }>\text { mass of mouse }  Therefore, effect of impact by elephant will be more than that of mouse . An elephant breaking into you will take you back faster than a mouse in space hits you.

8 0
3 years ago
A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f
zhuklara [117]

Answer:

5.791244495 KNm

Explanation:

The height h is given by, h=42.6sin42.3^{o}

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2} and h is already given hence substituting 77 Kg for m we obtain

PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm

PE=21.6567095 KNm

We also know that Kinetic energy is given by0.5mv^{2} where v is the velocity and substituting v for 20.3 we obtain

KE=0.5*77*20.3^{2}=15865.465 Nm

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

8 0
3 years ago
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