Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 × m
diameter d = 0.074 cm = 74 × m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W
Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
Therefore, the additional work needed is
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Answer:
a
The x- and y-components of the total force exerted is
b
The magnitude of the force is
The direction of the force is
Clockwise from x-axis
Explanation:
From the question we are told that
The magnitude of the first charge is
The magnitude of the second charge is
The position of the second charge from the first one is
The magnitude of the third charge is
The position of the third charge from the first one is
The position of the third charge from the second one is
The force acting on the third charge due to the first and second charge is mathematically represented as
Substituting values
The magnitude of is mathematically evaluated as
The direction is obtained as