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Mashutka [201]
4 years ago
15

A spherical raindrop 2.5 mm in diameter falls through a vertical distance of 3900 m. Take the cross-sectional area of a raindrop

= πr2, drag coefficient = 0.45, density of water to be 1000 kg/m3, and density of air to be 1.2 kg/m3. (a) Calculate the speed (in m/s) a spherical raindrop would achieve falling from 3900 m in the absence of air drag. m/s
Physics
1 answer:
Karolina [17]4 years ago
5 0

Answer:

276.62 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

Equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 3900+0^2}\\\Rightarrow v=276.62\ m/s

<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s

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As we know that in cellular phone all the signals are transmitted by antenna in the form of electromagnetic waves

All these electromagnetic waves are transmitted through the phones and then received as a signal to the receiver of another phone

These signals are then processed and converted into sound which is produced by the phone

So here correct answer will be

<em>C. The cellular telephone transmits information by electromagnetic waves to a receiver which then encodes them and produces sound.</em>

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3 years ago
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A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from res
Bumek [7]

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

t_1 = \sqrt{\frac{2H}{g}}

now the sound will come back to the observer in the time

t_2 = \frac{H}{v}

so we will have

t_1 + t_2 = 2.42

\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42

so we have

\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42

solve above equation for H

H = 26.8 m

Part b)

If sound reflection part is ignored then in that case

H = \frac{1}{2}gt^2

H = \frac{1}{2}(9.81)(2.42^2)

H = 28.7 m

so here percentage error in height calculation is given as

percentage = \frac{28.7 - 26.8}{26.8} \times 100

percentage = 7.18

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3 years ago
The heat pump is designed to move heat. This is only possible if certain relationships between the heats and temperatures at the
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4 years ago
A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential
ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

Explanation:

Given that,

Speed v= 3.50\times10^{5}\ m/s

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

U=qV=eV

Using conservation of energy

K_{i}+U_{i}=K_{f}+U_{f}

\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}

]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})

\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V

\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}

Put the value into the formula

\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}

\Delta V=639.2=0.639\ kV

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

\Delta U=q\Delta V

Put the value into the formula

\Delta U=1.6\times10^{-19}\times639.2

\Delta U=1.022\times10^{-16}\ J

(c). We need to calculate the work done on the proton

Using formula of work done

\Delta U=-W

W=q(V_{2}-V_{1})

W=-1.6\times10^{-19}(150-100)

W=-8\times10^{-18}\ J

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

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