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Ilia_Sergeevich [38]
2 years ago
7

A rubbit gets down from a rump which its /\x=0.85m in 0.5s, The rubbit's mass is 2kg, what is the net Force?

Physics
1 answer:
mestny [16]2 years ago
8 0

Answer: 13.6 N

Explanation:

The equation of motion for the rabbit is:

\Delta x=V_{ox}t+\frac{1}{2}a_{x}t^{2} (1)

Where:

\Delta x=0.85 m is the distance traveled by the rabbit

V_{ox}=0 m/s is the rabbit's initial velocity, assuming it started from rest

t=0.5 s is the time

a_{x} is the acceleration

Isolating a_{x}:

a_{x}=\frac{2 \Delta x}{t^{2}} (2)

a_{x}=\frac{2 (0.85 m)}{(0.5)^{2}} (3)

a_{x}=6.8 m/s^{2} (4)

On the other hand, the force F_{x} is given by:

F_{x}=m.a_{x} (5)

Where m=2 kg is the mass of the rabbit

Substituting (4) in (5):

F_{x}=(2 kg)(6.8 m/s^{2}) (6)

Finally:

F_{x}=13.6 N

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B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

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Plug in the values and solve for Pb

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Pd = mgh÷Δt

Given values:

m = 250.0kg

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h = 1.25m

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Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

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Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

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<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

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