The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W
Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º
Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-
)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
Answer:
Explanation:
Volume of the insulating shell is,

Charge density of the shell is,

Here, 

B)
The electric field is 
For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.
C)
For R <r <2R According to gauss law

substitute 

D)
The net charge enclosed for each r in this range is positive and the electric field is outward
E)
For r>2R
Charge enclosed is zero, so electric field is zero
Explanation:
Work done is the force applied to move a body through a specific or particular direction.
It is also the difference in the amount of energy expended in using an effort.
Work done is given as;
Work done = F x d CosФ
F is the force applied
d is the displacement
Ф is the angle
The unit of work done is in Joules.