The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
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A
More concentrated means more collisions per unit volume, and as the volume stays the same and only concentration changes, the there are more collisions
Answer:
6.67×10¯⁹ A
Explanation:
From the question given above, the following data were obtained:
Quantity of electricity (Q) = 2 μC
Time (t) = 5 mins
Current (I) =?
Next, we shall convert 2 μC to C. This can be obtained as follow:
1 μC = 1×10¯⁶ C
Therefore,
2 μC = 2 μC × 1×10¯⁶ C / 1 μC
2 μC = 2×10¯⁶ C
Next, we shall convert 5 mins to seconds. This can be obtained as follow:
1 min = 60 secs
Therefore,
5 min = 5 min × 60 sec / 1 min
5 mins = 300 s
Finally, we shall determine the current in the circuit. This can be obtained as follow:
Quantity of electricity (Q) = 2×10¯⁶ C
Time (t) = 300 s
Current (I) =?
Q = It
2×10¯⁶ = I × 300
Divide both side by 300
I = 2×10¯⁶ / 300
I = 6.67×10¯⁹ A
Thus, the current in the circuit is 6.67×10¯⁹ A
