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Drupady [299]
3 years ago
9

Heather and Jerry are standing on a bridge 60 m above a river Heather throws a rock straight down with a speed of 14 m/s. Jerry,

at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance.
a. How much time elapses between the first splash and the second splash?
b. Which rock has the faster speed as it hits the water?
Physics
1 answer:
melisa1 [442]3 years ago
4 0

Answer:

Time elapsed   = 2.856

SECOND ROCK HAS FASTER SPEED

Explanation:

This can be achieved in 2 ways, and it is clear from the careful study that the travel time is attributed to the time required to cross the path above the bridge by the 2nd stone  i.e.

14 = 9.8 * t

t = 1.428 sec and

time between 2 splash  = 2 * 1.428 = 2. 857 sec

2nd way

let time for 1st spalsh be t and

second splash be t1.

from equation of motion

so 60 = 14t + 0.5 * 9.8 * t * t

4.9 t^2 + 14t -60 = 0

On solving, we get t = 2.351 sec

for t_1, we have some extra time, which can be divided into t2 and t3.

14 = 9.8 * t2

t2 = 1.428 sec  

total time taken by 2 is

T = 1.428 + 1.428 + 2.351

so we get t_1 = 5.207 sec

time elapsed is  = T - t

                           = 5.207 - 2.351

Time elapsed   = 2.856

SECOND ROCK HAS FASTER SPEED

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To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

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Answer:

Part(a): The capacitance per unit length of the capacitor is \bf{1.56 \times 10^{-10}}~F~m^{-1}.

Part(b): The charge in the inner shell is + 1.6 \times 10^{-10}~C and the charge of the outer shell is - 1.6 \times 10^{-10}~C.

Explanation:

Part(a):

The capacitance (C) per unit length of a cylindrical capacitor is given by

C = \dfrac{2 \pi \epsilon_{0}}{log(b/a)}

where '\epsilon_{0}' is the permittivity of free space, 'b' is the radius of the outer shell and 'a' is the radius of the inner shell.

Given, b = 3.4~mm = 0.0034~m and a = 1.5~mm = 0.0015~m. We know, \epsilon_{0} = 8.854 \times 10^{-12}~F~m^{-1}. Substituting the values in the above equation,

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Part(b):

As the voltage on the inner conductor is higher than that of the outer conductor, positive charge resides on the inner shell and negative charge resides on the outer shell.

The charge 'Q' of a capacitor of length 'L' having capacitance 'C' and potential difference 'V' is given by

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Given, V = 350~mV = 350 \times 10^{-3}~V and L = 3.0~m. Substituting these values in the above expression

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The charge in the inner shell is + 1.6 \times 10^{-10}~C and the charge of the outer shell is - 1.6 \times 10^{-10}~C.

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