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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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If an object undergoes a change in momentum of 10 kg m/s in 3 s ,then the force acting on it is
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A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
3 years ago
PLEASE HELP. NOTICE HOW THEY PUT FUTURE
MrMuchimi

I think its <u>B</u> because it looks like it might indicate a future rain storm.

6 0
3 years ago
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