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Tresset [83]
1 year ago
5

A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin

etic friction between the crate and the floor?
Physics
1 answer:
____ [38]1 year ago
6 0

Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.

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suppose that you had an ohmmeter and a mystery component that could be either an inductor or a capacitor. How could you use the
sukhopar [10]

Answer:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

if the resistance > 1( very high )   then component is a capacitor

if the resistance = 0 then component is an inductor

Explanation:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

if the resistance > 1( very high )   then component is a capacitor

if the resistance = 0 then component is an inductor

8 0
3 years ago
A 2-kg book falls off a shelf. it hits a student traveling 2 m/s. how much kinetic energy does the book have?
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5 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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