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polet [3.4K]
2 years ago
8

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing

the lighting and hearing the sound was 5 second. Calculate the distance of the colliding cloud from the observer.
Physics
1 answer:
hodyreva [135]2 years ago
8 0

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

d=v\times t

d=340\ m/s\times 5\ s  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

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If the applied force is in the same direction as the object's displacement, the work done on the object is:

W = Fd

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Given values:

F = 45N

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Plug in and solve for W:

W = 45(12)

W = 540J

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A small plastic bead has been charged to -15nC.
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Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

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(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

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E = 1.67 x 10^6 N/C

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(D) the direction of acceleration is away from the bead, as the force is repulsive.  

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3 years ago
A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf
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(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

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The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

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F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

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P=Fv=(4829)(2.20)=10624 W

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The change in potential energy of the car is:

\Delta U = mg \Delta h

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g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

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