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diamong [38]
3 years ago
6

A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b

ottom of the cliff? (b) What is its speed just before hitting? And (c) what total distance did it travel? (Ans: 5.2 sec, 38.94 m/s, 84.7 m)
Physics
1 answer:
jonny [76]3 years ago
8 0

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0  

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

   v = u + a t

   v = 12 - 9.8 x 5.2

  v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

  vertical distance covered by the stone

 v² = u² + 2 g h

 0 = 12² - 2 x 9.8 x h

 h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

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Answer:

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Explanation:

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Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = \frac{3}{2}kT

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

K.E. = W

\frac{3}{2}kT = qV

Rearrange the above equation in terms of T :

T= \frac{2qV}{3k}

Substitute the suitable values in the above equation.

T=\frac{2\times1.6\times10^{-19}\times10.3 }{3\times1.38\times10^{-23} }

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3 years ago
A simple series circuit consists of a 150 Ω resistor, a 29 V battery, a switch, and a 2.1 pF parallel-plate capacitor (initially
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Find the electric flux and the disp at t=0.50ns 
<span>Given: </span>
<span>Resistor R = 160 Ω </span>
<span>Voltage ε = 22.0 V </span>
<span>Capacitor C = 3.10 pF = 3.10 * 10^-12 F </span>
<span>time t = 0.5 ns = 0.5 * 10^-9 s </span>
<span>ε0 = 8.85 * 10^-12 </span>
<span>Solution: </span>
<span>ELECTRIC FLUX: </span>
<span>Φ = Q/ε0 </span>
<span>we have ε0, we need to find Q the charge </span>
<span>STEP 1: FIND Q </span>
<span>Q = C ε ( 1 - e^(-t/RC) ) </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } </span>
<span>Q = 43.31 * 10^-12 C </span>
<span>STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> </span>
<span>Φ = Q/ε0 </span>
<span>Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } </span>
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<span>DISPLACEMENT CURRENT </span>
<span>we use the following equation: </span>
<span>I = { ε / R } { e^(-t/RC) } </span>
<span>I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>I = { 0.1375 } { 0.365 } </span>
<span>I = 0.0502 A = 0.05 A </span>
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Answer:

B

Explanation:

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I think it is neutral
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Answer:

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Explanation:

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In this case we know that the person walks from home to work and then walks from work to home. Therefore, the total displacement is the linear distance between the point where its journey begins and the point where the route ends.

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