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user100 [1]
4 years ago
13

Forces that are equal in strength and opposite indirection are

Physics
2 answers:
blsea [12.9K]4 years ago
8 0
They are called balanced forces.
Otrada [13]4 years ago
4 0
Hey there! My name is Christy and I'm gladly to help you out!


The three main forces that stop moving objects are friction, gravity and wind resistance. Equal forces acting inopposite directions are called balanced forces. Balanced forcesacting on an object will not change the object's motion. When you add equal forces in opposite direction, the netforce is zero.

Hope this helped!


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A chemotroph gains its energy by eating something else
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You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongs
nekit [7.7K]

There is one mistake in the question as unit of height of building is not given.So I assume it as meter.The complete question is here

You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head, where should the professor be when you release the egg? Assume that the egg is in free fall.  

Answer:

d=3.67 m

Explanation:

Height of building=46.0 m

First we need to find time taken by egg to reach 1.80 m above the surface

So to find time use below equation

S=vt+\frac{1}{2} gt^{2}\\ (46.0-1.80)m=(om/s)t+\frac{1}{2}(9.8m/s^{2} )t^{2}\\t=\sqrt{\frac{(46.0-1.80)m}{4.9} }\\ t=3.06s

As velocity 1.20m/s is given and we have find time.So we can easily find the distance

So

distance=velocity*time\\d=v*t\\d=(1.20m/s)*(3.06s)\\d=3.67m

3 0
3 years ago
A ball of moist clay falls 17.0 m to the ground. It is in contact with the ground for 19.0 ms before stopping,
Montano1993 [528]

Explanation:

It is given that,

Initial height of the ball, x_o=17\ m

As it stops, its position, x = 0

It is in contact with the ground for 19.0 ms before stopping, t=19\times 10^{-3}\ s

Initial speed of the ball, u = 0

(a) Let a is the acceleration of the ball during the time it is in contact with the ground. Its can be calculated suing equation of kinematics as :

v^2-u^2=2a(x-x_o)

a =-g

v^2=-2g(x-x_o)

v^2=-2g(-17)

v^2=-2\times 9.8\times (-17)    

v = 18.25 m/s

(b) The average acceleration can be calculated as :

a_{avg}=\dfrac{v-u}{t}

a_{avg}=\dfrac{18.25-0}{19\times 10^{-3}}

a_{avg}=960.52\ m/s^2

As the value of acceleration is positive, so the ball is accelerating in upward direction.

4 0
3 years ago
Equilibrium: A 10.0-kg picture is held in place by two wires, the first one hanging at 50.0° to the left of the vertical and the
Sonja [21]

Answer:

The tension in the first wire is 70.24N and the tension in the second wire is 75.36N

Explanation:

Check attachment for solution

6 0
3 years ago
I NEED THIS BY TODAY DDD:
MAVERICK [17]

Answer:

I can't see the picture, can you right down the same thing that is the picture in the comments so I can help you.

Explanation:

Thx.

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