Answer:
P=627.47W
Explanation:
To solve this problem we have to take into account, that the work done by the winch is

the force, at least must equal the gravitational force

with force the tension in the cable makes the winch go up.
The work done is

To calculate the power we need to know what is the time t. But first we have to compute the acceleration
The acceleration will be

and the time t

The power will be

HOPE THIS HELPS!!
Answer:
T = g μ_s ( M+m )
78.4 N
Explanation:
When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .
Net force on the lower box
= T - mg μ_s = M a ( a is the acceleration created by net force in M )
Considering force on the upper box
mg μ_s = ma
a = g μ_s
Put this value of a in the equation above
T - m gμ_s = M g μ_s
T = mg μ_s + M g μ_s
= g μ_s ( M+m )
2 )
Largest tension required
T = 9.8 x .50 x ( 10+6 )
= 78.4 N
Answer:
heya answer option b
Explanation:
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Answer:
457.81 Hz
Explanation:
From the question, it is stated that it is a question under Doppler effect.
As a result, we use this form
fo = (c + vo) / (c - vs) × fs
fo = observed frequency by observer =?
c = speed of sound = 332 m/s
vo = velocity of observer relative to source = 45 m/s
vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).
fs = frequency of sound wave by source = 459 Hz
By substituting the the values to the equation, we have
fo = (332 + 45) / (332 - (-46)) × 459
fo = (377/ 332 + 46) × 459
fo = (377/ 378) × 459
fo = 0.9974 × 459
fo = 457.81 Hz
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as







now the electric field is given as


Part b)
Now since the electric field is required at same distance on other side
so the field will remain same on other side of the plate
