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hichkok12 [17]
3 years ago
5

An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5

Physics
1 answer:
dexar [7]3 years ago
5 0

Answer:

The alligator will take t = 10 s to reach the final speed of 35 m/s

Explanation:

As we know that the initial speed of the alligator is 5 m/s

then it accelerate by given acceleration to reach the final speed of 35 m/s

so we will have

v_i = 5 m/s

v_f = 35 m/s

a = 3m/s^2

now we have

v_f = v_i + at

35 = 5 + 3 t

t = 10 s

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State the principle of floatation​
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<em><u>Principle of Floatation</u></em>

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A trampoline has a spring constant of 3430 N/m how far will the trampoline sink when a 70 kg person steps on it
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True or false. Atoms of a given element will have the same mass.
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True

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4 0
3 years ago
A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =
Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = 0.5\times 10^{-4} kg

Weight of the rain drop = W

Duration of time = t = 3 seconds

W=m\times g

Drag force on rain drop = D=0.2\times 10^{-5} v^2

W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N

Motion of the rain drop:

F=m\times a

Net force on the rain drop , F=  W - D

W-D=m\times a

0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a

0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}

0.006v^2+0.05v-1.5=0

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

12.18 m/s=0m/s+a\times 3 s

a=4.06 m/s^2

s=ut+\frac{1}{2}at^2 (second equation of motion)

s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0
2 years ago
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