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3 years ago

Please please please please help

Physics

2 answers:

aivan3 [116]3 years ago

4 0

Answer:

I am pretty sure its the second one but I could be wrong sorry if I am.

Explanation:

Nikitich [7]3 years ago

3 0

Second one i think yeah

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A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

2 years ago

If I am throwing a bowling ball off a roof and a watermelon off a roof at 30

rosijanka [135]

Answer:

Th ball

Explanation:

8 0

3 years ago

Why will the circuit stop working if the wites were made out of plastic instead of copper?

anastassius [24]

Answer:

plastic isn't conductive

Explanation:

Correct me if I'm wrong but plastic isn't conductive copper is so if it's not conductive it will probably mess up the circuit or somthing

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2 years ago

What is the direction of the net force that acts on an object undergoing uniform circular motion?

valentinak56 [21]

Answer B. the direction of force is the same

7 0

3 years ago

The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. A) If the car’

ladessa [460]

Answer:

Acceleration is -30000 mi/h²

Distance travelled in the 3 seconds of deceleration is 261.888 feet

Explanation:

t = Time taken for the car to slow down = 3 s =

u = Initial velocity = 75 mi/h

v = Final velocity = 50 mi/h

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{50-75}{\frac{3}{3600}}\\\Rightarrow a=-30000\ mi/h^2$

Acceleration is -30000 mi/h²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{50^2-75^2}{2\times -30000}\\\Rightarrow s=0.0496\ mi$

Converting to feet

1 mile = 5280 feet

0.0496 mile = 0.0496×5280 = 261.888 feet

Distance travelled in the 3 seconds of deceleration is 261.888 feet

7 0

3 years ago