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Vanyuwa [196]
3 years ago
14

A boy is pulling his two sisters on a sled.

Physics
1 answer:
OlgaM077 [116]3 years ago
4 0

Answer:

Explanation:

The oly way we can figure this out is if the boy is pulling the sled at a constant velocity. If not, we need a value for acceleration, and you don't have that here. If the boy is pulling the sled at a constant velocity, then the value for acceleration is 0, making this a really simple problem. I'm going with that, since there is no way to answer you otherwise. If velocity is not constant, please either repost the question or put it in the notes section under the question as it stands. If acceleration is 0, then

F - f = ma becomes

F - f = m(0) which is

F - f = 0 and

F = f which says that the applied force is the same as the frictional force. We need then to find the frictional force, which has an equation of

f = μF_n where normal force is the same as the weight of the 2 girls. We will find that, then:

Each girl's mass is different so the normal force/weight equation is

w = (30.0)(9.8) + (40.0)(9.8) to get

w = 290 + 390 and

w = 680. Plug that into the frictional force equation:

f = (.120)(680) so

f = 82N

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What is E(r)E(r)E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the dist
andriy [413]

Answer:

E(r) = λ/2πrε0

Explanation:

If we consider an infinitely long line of charge with the charge per unit length being λ, we can take advantage of the cylindrical symmetry of this situation.

By symmetry, i mean that the electric fields all point radially away from the line of charge and thus there is no component parallel to the line of charge.

Niw, let's use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.

Doing that will mean that the electric field would be perpendicular to the curved surface of the cylinder. Therefore, the angle between the electric field and area vector is equal to zero and cos θ = cos 0 = 1

Now, the top and bottom surfaces of the cylinder will lie parallel to the electric field. Therefore, the angle between the area vector and the electric field would be 90° and cos θ = cos 90 = 0

Now, we know that according to Gauss Law,

Electric Flux, Φ = E•dA

Thus,

Total Φ = Φ_curved + Φ_top + Φ_bottom

Thus,

Φ = ∫E•dA cos 0 + ∫E•dA cos 90° + ∫E•dA cos 90°

We now have ;

Φ = ∫E . dA × 1

Since we are dealing with the radial component, the curved surface would be equidistant from the line of charge and the electric field in the surface will be the same magnitude throughout.

Thus,

Φ = ∫E•dA = E∫dA = E•2πrl

The net charge enclosed by the surface is given by:

q_net = λl

So using gauss theorem, we have;

Φ = E•2πrl = q_net/εo = λl/εo

E•2πrl = λl/ε0

Making E the subject, we obtain ;

E = λ/2πrε0

4 0
3 years ago
Jorge visits Panama city, Florida during the month of may. he feels a shore breeze blowing from the ocean on the beach. what cau
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