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LuckyWell [14K]
3 years ago
5

Two drums of the same size and same height are taken.

Physics
1 answer:
Gelneren [198K]3 years ago
3 0

Answer:

i) The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) The pressure acting on the base of <em>B</em> will be the same as the pressure acting on the base of <em>A</em>

iii) The pressure on the base of drum <em>A</em> will be slightly less than the pressure on the base of drum <em>B</em>

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

h_{max} = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If <em>A</em> is completely filled, we have h_A = h_{max}

Therefore, P_A = h_{max}×\rho_{liquid}×g

If <em>B</em> is half filled, we have, h_B =  (1/2)·h_{max}

P_B = (1/2) × h_{max}×\rho_{liquid}×g

Therefore, P_B = (1/2) × P_A

The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) If both <em>A</em> and <em>B</em> are each filled with water (the same liquid), then the pressure on their bases will be P_A = h_{max}×\rho_{water}×g = P_B, the same, given that the acceleration due to gravity, <em>g</em>, is constant and the same in Nepal and India

iii) If <em>A</em> is filled with water, and <em>B</em> is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

P_A = h_{max}×\rho_{water}×g <  P_B =

The pressure on the base of drum <em>A</em> will be less than the pressure on the base of drum <em>B.</em>

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Explanation:

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Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
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To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

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v is the speed of the object

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In the problem, the original centripetal acceleration is

a=0.50 m/s^2

We want to increase it by a factor of 4, i.e. to

a'=2.00 m/s^2

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

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a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

r'=\frac{1}{4}r

So the new acceleration is

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Learn more about centripetal acceleration:

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A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?
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Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

         Known:

  • R = 10.0 Ω                            
  • V = 12.0 V

       Unknown:

  • I = ???

I = V/R

= 12.0 V / 10.0 Ω

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While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r
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Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m

If f = 78 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

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Answer:

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We know that capacitance is given by C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F

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3 years ago
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