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Leona [35]
3 years ago
5

When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sl

iding, however, you can use a force of only 1.06 N to keep the book moving with constant speed. Part A What is the coefficient of static friction between the book and the tabletop?
Physics
1 answer:
zimovet [89]3 years ago
3 0
Static frictional force = ƒs = (Cs) • (Fɴ)
              2.26 = (Cs) • m • g
              2.26 = (Cs) • (1.85) • (9.8)
           Cs = 0.125 
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
              1.49 = (Cκ) • m • g
              1.49 = (Cκ) • (1.85) • (9.8)
           Cκ = 0.0822 
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kompoz [17]

Answer:

Instantaneous velocity

Explanation:

Acceleration = rate of change of velocity

Average velocity = total displacement divided by total time taken

Position = It shows the location of the object

Instantaneous velocity = It is the velocity of an object at a particular instant.

The term that describes how fast and in what direction an object is moving at a particular moment is "instantaneous velocity". It is basically equal to the derivative of position wrt time.

7 0
3 years ago
What is the relationship between force of gravity and mass
sweet [91]

Answer:

I hope the picture below help.

Explanation:

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3 years ago
A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What
Ratling [72]

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

<u>V₀y = 0 m/s</u>

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

<u>t = 2.47 s</u>

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

<u>V₀ₓ = 61.86 m/s</u>

<u></u>

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

4 0
3 years ago
A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×
otez555 [7]

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, m=1.01\times 10^7\ kg

Force applied on the tracks, F=7.5\times 10^5\ N

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s

So, the required time is 300.3 seconds.

4 0
3 years ago
We know that the motion of the Moon around the Earth is due
nadezda [96]

Answer:

YES

Explanation:

Gravity acts as the centripetal force and the velocity earth has keeps it from falling on the sun.

3 0
3 years ago
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