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2 years ago

A bicycle has a mass of 10kg and an acceleration of 2m/s². What is the net force of the bicycle?

Physics

1 answer:

kobusy [5.1K]2 years ago

4 0

Answer:

<h3>The answer is 20 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 10 kg

acceleration = 2 m/s²

We have

Force = 10 × 2

We have the final answer as

Hope this helps you

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Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

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3 years ago

Mercury is in the 80th position in the periodic table. How many protons does it have?

aalyn [17]

Mercury has 80 protons.

4 0

3 years ago

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students may elect to take the Knowledge Test online or at Drivers Office what additional test must be taken at the licensing of

Nutka1998 [239]

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3 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

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2 years ago

Suppose two objects are gravitationally attracted to each other with some force F. If the mass of object 1 is multiplied by a fa

RoseWind [281]

If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is

F' = G (5m₁) (2m₂) / d²

F' = 10 Gm₁m₂ / d²

but this is really just F' = 10F. So J is the correct choice.

8 0

2 years ago

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