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Illusion [34]
3 years ago
6

How do you find the velocity after a collision

Physics
1 answer:
Evgen [1.6K]3 years ago
7 0

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

Explanation:

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How would the period of pendulum differ from an equivalent one on earth?<br>​
Mars2501 [29]

Answer:

If you know that that free fall acceleration g on the Moon is about 6 times less than on the Earth, it gives you the answer: on the Moon the same pendulum will have a period about √6≈2.45 longer than on the Earth.

5 0
3 years ago
g A simple pendulum (consisting of a point mass suspended by a massless string) on the surface of the earth has a period of 1.00
leonid [27]

Answer:

Explanation:

The formula for time period of a pendulum is given as follows :

T = 2π\sqrt{\frac{l}{g} }

l is length of pendulum and g is acceleration due to gravity .

So time period of pendulum  is not dependent on the mass of the pendulum . If time period is same and length is also the same then acceleration due to gravity will also be the same . Hence the acceleration due to gravity at distant planet will be same as that on the earth.

5 0
2 years ago
What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest
Lerok [7]

|acceleration|  =  (change in speed) / (time for the change)

Change in the car's speed = (27 - 0) = 27 m/s
Time for the change  =  10 sec

|acceleration| = (27 m/s) / (10 s)  =  2.7 m/s² .

That's the magnitude of the car's acceleration.
We don't know anything about its direction.
5 0
2 years ago
(15 points) :^|
mihalych1998 [28]
That is FALSE. The equation to calculate the charges has a distance component that is in the denominator which means that it is inversely proportional (as the distance os greater the force is smaller)
7 0
3 years ago
A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly
Ratling [72]

Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

v = 8.854 \ m/s

The tension in the string is:

T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N

Thus, the   tension in the string an instant before it broke = 34 N

6 0
2 years ago
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