Answer:
Calculating Coefficient of friction is 0.229.
Force is 4.5 N that keep the block moving at a constant speed.
Explanation:
We know that speed expression is as
.
Where,
is initial speed, V is final speed, ∆s displacement and a acceleration.
Given that,
=3 m/s, V = 0 m/s, and ∆s = 2 m
Substitute the values in the above formula,

0 = 9 - 4a
4a = 9

is the acceleration.
Calculating Coefficient of friction:


Compare the above equation

Cancel "m" common term in both L.H.S and R.H.S





Hence coefficient of friction is 0.229.
calculating force:


F = 4.5 N
Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.
Answer:
factual evidence of customer-service levels.
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better understanding of the capability range of a process.
Answer:
initial magnetic field 1.306 T
Explanation:
We have given area of the conducting loop 
Emf induced = 1.2 volt
Initial magnetic field B = 0.3 T
Time dt = 0.087 sec
We know that induced emf is given by 


So initial magnetic field = 1.606-0.3= 1.306 T
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