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telo118 [61]
2 years ago
5

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect

ed at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing
Physics
1 answer:
AleksAgata [21]2 years ago
8 0

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

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Afina-wow [57]

First speed = 20km/h

Time = 3 hours

Distance = 3×20

<h3> = <u>60 km</u></h3>

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<h3> = <u>120 km</u></h3>

Total distance = 60+120 = <u>180km</u>

Total time = 3+4 =<u> 7 hours</u>

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Hope this will help...

4 0
3 years ago
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Ilia_Sergeevich [38]
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Why was the idea of plate tectonics difficult for many scientists to accept for many years after it was first introduced?
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3 0
3 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied
Lesechka [4]

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

KA = 2 x KB

Option A is correct

6 0
3 years ago
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