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Komok [63]
3 years ago
5

A5 kg box slides 3 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and t

he box if the box had an initial speed of 3 m / s?
Physics
1 answer:
frutty [35]3 years ago
6 0

Answer:

Coefficient of kinetic friction (Cof. KE) = 0.153

Explanation:

Given:

Mass of box (M) = 5 kg

Distance = 3 m

Initial speed (v) = 3 m/s

Find:

Coefficient of kinetic friction (Cof. KE)

Computation:

v² = u² + 2as

a = v² / 2s

a = 9 / 2(3)

a = 1.5 m/s²

Coefficient of kinetic friction (Cof. KE) = a / g

Coefficient of kinetic friction (Cof. KE) = 1.5 / 9.8

Coefficient of kinetic friction (Cof. KE) = 0.153

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Acceleration = 45m/s^2
Time = 2 s

Explanation:

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A block of cheese is sliding up an inclined surface that is not smooth.
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I cant fit the entire picture but this should help

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What is the difference between charging by contact and charging by induction in terms of electron transfer.
Veronika [31]

Answer:

the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch

Explanation:

There are three methods of charging an object:

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- Charging by contact: this is done by putting in contact a charged object with a neutral, conducting object. In this case, the charges are transferred from the charged object to the neutral object; at the end of the process, the neutral object will also have a net electric charge, so it will be also charged.

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3 years ago
Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a
Yanka [14]

Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ      .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ  = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)²     .....(2)

T Cosθ = mg    .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ =  0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

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