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3 years ago

A rower in a boat pushes the water using an oar.

Physics

2 answers:

brilliants [131]3 years ago

6 0

Answer:

It's B, because of process of elimination:

A: The rower would be applying force to the oar not the other way around. So that's immediately out.

C: If the force of the oar applies to the water, the force of the water cannot apply to the oar.

D: Same as why A is incorrect.

So B is the only one left.

solong [7]3 years ago

6 0

The correct answer is B. “Every action has an equal or opposite reaction”.

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sergejj [24]

Answer: $\sqrt{2}T$

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

$T'=2\pi \sqrt{\frac{2m}{k}}$

$T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}$

$T'=\sqrt{2}T$

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bija089 [108]

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2 years ago

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3 years ago

A rock is dropped from a height of 3.4 m. How much time does it take to hit

siniylev [52]

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3 years ago

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same

Hatshy [7]

Answer:

$v_2=\sqrt{2}v_1$

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

$v_1^{2} =v_0^{2} +2gh$

Since the object is initially at rest, v₁ becomes:

$v_1=\sqrt{2gh}$

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

$v_2^{2}=v_1^{2} +2gh$

Substituting v₁ in this expression and solving for v₂, we get:

$v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}$

Now, dividing v₂ over v₁, we get the expression:

$\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1$

It means that v₂ is √2 times v₁.

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3 years ago