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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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Bella’s average velocity is about 0.693 meters per second.

To find the average velocity, you must divide the distance by the change in time, which should look like v=d/t

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djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F

<h3>C1 and C2 are parallel to C3</h3>

C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F

<h3>C(123) is series to C5 and C6</h3>

\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F

<h3>C7 and C8 are in series</h3>

\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

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