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vlabodo [156]
3 years ago
5

A girl pulls her younger brother on a sled along a flat sidewalk (the total mass of the sled is 30kg). A frictional force of 50N

opposes the pull. When moving, the girl pulls on the sled with a force of 150N at an angle of 30 degrees.
How much time will it take for the sled to travel 50?

What is the normal force?
Physics
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

6.13 s

219 N

Explanation:

Newton's law in the x direction:

∑F = ma

150 cos 30° N − 50 N = (30 kg) a

a = 2.66 m/s²

Δx = v₀ t + ½ at²

(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²

t = 6.13 s

Newton's law in the y direction:

∑F = ma

Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0

Fn = 219 N

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Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

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In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

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