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almond37 [142]
2 years ago
5

A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.

what is the magnitude of the sum of the frictional forces acting on the bike and its rider?
Physics
1 answer:
mr_godi [17]2 years ago
4 0

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

brainly.com/question/1714663

#SPJ1

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What is the magnitude and direction (right or left) of the
Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

Fy_{net}=F_{n}+F_{g} (1)

Where F_{n}=12 N is the Normal force, directed upwards and F_{g}=-12 N is the weight of the box (gravity force), directed downwards.

Fy_{net}=12 N-12 N (2)

Fy_{net}=0 N (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

Fx_{net}=F_{left}+F_{right} (4)

Where F_{left}=-3 N and F_{right}= 15 N

Fx_{net}=-3 N + 15 N (5)

Fx_{net}=12 N (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0
3 years ago
A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi
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Answer:

Explanation:

a ) Time period  T = 2 s

Angular velocity ω = 2π / T

=  2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ =  I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

w_2=\frac{356.25\times3.14}{256.25}

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² -  1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

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3 years ago
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