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Mnenie [13.5K]
3 years ago
7

Work is done on an object when an applied force causes the object to move in the same direction as the force.

Physics
1 answer:
Lelu [443]3 years ago
3 0

Answer:yes

Explanation:

Work is done on an object when an applied force causes the object to move in the same direction as the force

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d. nothing

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a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in
Lina20 [59]

Answer: 0.5 m

Explanation:

Given

Mass of the person is m=50\ kg

Trampoline launches the person into the air up to height of h=2\ m

Force experience by springs is F=1960\ N

Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.

\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m

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Why do the planets appear in different locations in the night sky while the pattern of stars in a constellation stays the same?
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The nearest star outside the solar system is almost 32 thousand times
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This all makes me feel small.  How about you ?
7 0
3 years ago
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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit
DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

So here we can say that the spear will end up above the fish relative to the actual position of the fish

5 0
3 years ago
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