E = hf = hc/λ
= (6.6 x 10^-34) × (3 × 10^8) ÷ (3.1 × 10^-7)
=6.138 × 10^-32 J
a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Answer:
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Answer:
Power in the circuit is 0.1 amp
Explanation:
The power in the circuit is given by the formula
P = V x I
Where P is Power, V is voltage supplied and I is current in circuit.
so, I = P/v
= 2/20
=0.1 A
Study more about power
<u>https://brainly.in/question/1063947</u>
Answer:
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Explanation:
From the question we are told that:
Height 
Radius 
Height of water 
Gravity 
Density of water 
Generally the equation for Volume of water is mathematically given by
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
Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by
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Substituting dv


Therefore
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
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
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