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zavuch27 [327]
1 year ago
15

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

without sinking below the surface of the water? (Note that water has a density of 1.00 g/cm³.)
Physics
1 answer:
Doss [256]1 year ago
3 0

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}}\!.

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4 0
3 years ago
The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin
Ganezh [65]

Answer:

The current in the circuit decrease slowly .

Explanation:

Given as :

For the electrical circuit

The voltage V in the circuit is slowly decreasing

The resistance R of the resistor slowly increasing after heating

Now, From Ohm's Law

Voltage is directly proportional to the flow of current through circuit

I.e V ∝ I

Or. V = R × I

where R is the proportionate constant and this is the resistance of the resistor

whose property is to opposes the flow of current in the circuit

So, If R value more then current I reduces in the circuit

∵ Here in the circuit ,  The resistance is slowly increasing, so, current I is slowly decreasing .

Hence The current in the circuit decrease slowly . answer

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3 years ago
What unita are used to measure electrical current​
JulijaS [17]

Answer:

It's called an ampere!

Explanation:

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Hope this helps :)

4 0
3 years ago
Read 2 more answers
An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

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3 years ago
What burns quicker, white or colored candles?
VashaNatasha [74]
White is a hotter color so it burns faster. :)
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