Answer:
(a) coefficient of friction = 0.451
This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)
(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.
Explanation:
The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.
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Answer:
a)
m/s
b)
Angular frequency =
Explanation:
As we know
q is the charge on the electron = C
B is the magnetic field in Tesla = T
r is the radius of the circle = m
mass of the electrons = Kg
a)
Substituting the given values in above equation, we get -
m/s
b)
Angular frequency =
Answer:
South = 1.5m
West =4.2m
Explanation:
Kindly see attached a rough draft of the situation
Step one
Given data
From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse
Step two:
Hence for an opponent to tackle him towards the south, he must be at
sin θ= opp/hyp
sin 20=x/4.5
x=sin 20*4.5
x=0.342*4.5
x= 1.5m
Also, for an opponent to tackle him towards the south, he must be at
cos θ= adj/hyp
cos 20=y/4.5
y=cos 20*4.5
y=0.93*4.5
y= 4.2m
