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dusya [7]
3 years ago
14

How can an object overcome static friction?

Physics
1 answer:
larisa86 [58]3 years ago
4 0

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

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1. A man throws a ball up with a velocity of 30 m/s. How high does it get?
ElenaW [278]
Yes the answer is g=5
8 0
3 years ago
A ball that is thrown upwards from the ground will eventually reach its highest point and fall back to the ground. Which of the
tia_tia [17]

Answer:

option (c)

Explanation:

When an object thrown upwards, the value of acceleration acting on the object is acceleration due to gravity which is always acting towards the earth.

As it falls downwards, the acceleration is again equal to the acceleration due to gravity.

So, the ball's acceleration is constant.

5 0
3 years ago
An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec
Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0
3 years ago
The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pu
Kisachek [45]

Answer:

a=2.36\ m/s^2

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass  less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

m_1g-m_2g=(m_1+m_2)a

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

a=2.36\ m/s^2

Lets tension in string is T

m_1g-T=m_1a

T=m_1(g-a)

T=21.1(9.81-2.36) N

T=157.06 N

6 0
3 years ago
An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is
Lemur [1.5K]

Answer:

      R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

           R = v₀² sin  2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

           R = 9.14² sin (2  35) / 9.8

           R = 8.01 m

this is the jump length

5 0
3 years ago
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