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ELEN [110]
2 years ago
11

Hey can anyone please help me with this it’s due in few hours and I’m stuck

Physics
1 answer:
Margaret [11]2 years ago
7 0

Answer:

1)  P = rho g h = d g h    where d is replacing the greek letter rho (density)

a)   pressure at A must be 1/2 that of B because of 1/2 the height

b) pressure at B equals that at D because the height is the same

 (Pascal's principle - pressure in a fluid is same in all directions)

c) pressure A is the same as that C

d) the pressure at B would be .8  of the water at B if paraffin replaced the water  (P = d g h = .8 * dw g h)      due to change in density

2)   Volume = 4 m X 3 m X 2 m) = 24 m^3    volume of tank

b) mass = volume * density  = 24 m^3 * 800 kg / m^3 = 19,200 kg

c) weight = mass * gravity = 19,200 kg * 9.8 m/s^2 = 188,160 Newton's

d) P = d g h = 800 kg/m^3 * 9.8 m/s^2 * 2 m 15,680 N/m^2

Pressure is given in force/area or N/m^2

3) Pw / Pp = dw / dp = 1000 / 800 = 1.25

Pw = 1.25 Pp = 15,680 m/s^2 * 1.25 = 19,600 Pascal's

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Original time period of the pendulum (T) = 3.45 s

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New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

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So, for the new length, the time period is given as:

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Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

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