The period of the tan function is π so (∅ + π) would yield the same value as ∅
F(∅ + π) = 3
The easiest way to answer this question is just to get the answer first. The answer is A with the added comment that no chemical reaction has taken place.
Layered means that the chemicals are not soluble in one another. B is not the answer.
C is eliminated by what what was said about A.
D a solution is not a pure substance (singular) by itself. There are at least 2 chemicals together.
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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Answer:
25000 V
Explanation:
The formula for potential is
V = Kq/r
Potential at B due to the charge placed at origin O
V1 = K q / OB

V1 = 10000 V
Potential at B due to the charge placed at A
V2 = K q / AB

V2 = 15000 V
Total potential at B
V = V1 + V2 = 10000 + 15000 = 25000 V
In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.
Part a)
when we pull the rod with constant speed then power required will be product of force and velocity
here we will have

P = 4 W
v = 4 m/s
now we will have


So external force required will be 1 N
PART B)
now in order to find magnetic field strength we can say

here we know that induced EMF in the wire is E = vBL
so power due to induced magnetic field is given by


by solving above equation we will have
