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BabaBlast [244]
2 years ago
14

Make a diagram showing the forces acting on a coasting bike rider traveling at 25km/h on a flat roadway.

Physics
2 answers:
Vikentia [17]2 years ago
5 0

Answer: see the figure with the diagram attached.

Explanation:

1) The diagrams showing the forces acting on an object (e.g. a coasting rider) are called free body diagrams.

2) These diagrams consider the object a point, remove all the contacts of the object, and replace all the interactions with force vectors.

3) In the given case, these are the elements that need to be included in the digram:

i) object: the bike rider with the bike. It is represented with a solid dot.

ii) Interaction between the center of the Earth and the object.

A vertical force (vector) pointing downward, whose magnitude equals the weight of the rider and the bike: Weight = mass × g.

It is designated by W.

iii) Interaction betweeen the object and the ground.

A vertical force (vector) pointing upward. It is the reaction force of the ground on the rider/bike. It is responsible for the rider/bike do not fall down toward the Earth.

It is called normal force and designated by N

iv) Force forward.

It is the force that produces the motion forward. It is paralled to the ground (horizontal).

It is directed to the right and desingnated by Ff

v) Resistance and drag force.

Are forces that oppose the motion forward (air resistance and friction with the pavement).

Since, the velocity is told to be constant, the acceleration is zero, so the net force acting is also zero.

That will lead to the fact that the forward motion is equal to the friction and drag forces).

it is horizontal, directed to the left and designated Fr.

Now you can watch and understand the diagram attached.

4) From the diagram and the balance of forces (constant velocity = zero acceleration = zero net force), you can ge the equations of the motion:

Ff = Fr

N = W

valkas [14]2 years ago
3 0
<em>[see the picture attached]

The forces are in red.</em>

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q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C

The values of the second charge is 1.64 μC

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