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3 years ago

The magnitude of the frictional force along the plane is most nearly

Physics

1 answer:

Svetach [21]3 years ago

3 0

Answer:

F = N*μ or F =m*g*μ

Explanation:

The friction force is defined as the product of the normal force by the corresponding friction factor.

When a body is in equilibrium over a horizontal plane its normal force value shall be equal to:

$N = m*g\\where:\\m=mass [kg]\\g=gravity [m/s^2]\\N= normal force [N]$

if we simplify this formula more for a balanced body on a horizontal plane, we will have.

$F=m*g*u$

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3 years ago

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A 70 kg object strikes the ground with 2500 J of KE after falling freely from rest. How far above the ground was the object when

Kitty [74]

Answer:

3.64 m

Explanation:

m = Mass of object = 70 kg

Kinetic energy of the object = 2500 J

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

h = Height from which the object is dropped

Kinetic energy is given by

$\dfrac{1}{2}mv^2=2500\\\Rightarrow v^2=\dfrac{2\times 2500}{70}$

From conservation of energy we get kinetic energy equal to potential energy.

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{1}{2g}v^2\\\Rightarrow h=\dfrac{1}{2\times 9.81}\times \dfrac{2\times 2500}{70}\\\Rightarrow h=3.64\ \text{m}$

The object was released from a height of 3.64 m.

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3 years ago

What is the average velocity in the time interval 5 to 6 seconds?

Alexeev081 [22]

Answer:

The average velocity from 5 to 6 seconds is -27 m/s

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3 years ago

You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0

stellarik [79]

Answer:

Vi = 24.14 m/s

Explanation:

If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

Loss in K.E of the Rock = Gain in P.E of the Rock

(1/2)(m)(Vi² - Vf²) = mgh

Vi² - Vf² = 2gh

Vi² = Vf² + 2gh

Vi = √(Vf² + 2gh)

where,

Vi = Rock's Speed as it left the ground = ?

Vf = Final Speed = 17 m/s

g = 9.8 m/s²

h = height of rock = 15 m

Therefore,

Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]

Vi = √583 m²/s²

Vi = 24.14 m/s

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4 years ago

The mass and radius of Earth are____ME and____RE, respectively. The mass and radius of the Moon are approximately 186ME and 27RE

oksano4ka [1.4K]

Answer:

1. Mass of the Earth is 5.97219 × 10^24 kg

2. Radius of the Earth is 6,371 km

3. Acceleration due to gravity on the surface of the Earth is 9.81 m/s^2

4. Acceleration due to gravity on the surface of the moon is 1.625 m/s^2

Explanation:

The mass and radius of Earth are 5.97219 × 10^24 kg ME and 6,371 km RE, respectively. The mass and radius of the Moon are approximately 186ME and 27RE, respectively. The acceleration due to gravity on the surface of Earth is 9.81 m/s^2 . The acceleration due to gravity on the surface of the Moon is most nearly 1.625 m/s^2

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3 years ago

The magnitude of the frictional force along the plane is most nearly​

The magnitude of the frictional force along the plane is most nearly