Answer:
Total energy saving will be 0.8 KWH
Explanation:
We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW
30 bulbs are of power 60 W
So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW
Total power of 80 bulbs = 1.8+5 = 6.8 KW
Total time = 3 hour
We know that energy 
Now power of each CFL bulb = 25 W
So power of 80 bulbs = 80×25 = 2000 W = 2 KW
Energy of 80 bulbs = 2×3 = 6 KWH
So total energy saving = 6.8-6 = 0.8 KWH
Answer:
(a) 0.05 Am^2
(b) 1.85 x 10^-3 Nm
Explanation:
width, w = 10 cm = 0.1 m
length, l = 20 cm = 0.2 m
Current, i = 2.5 A
Magnetic field, B = 0.037 T
(A) Magnetic moment, M = i x A
Where, A be the area of loop
M = 2.5 x 0.1 x 0.2 = 0.05 Am^2
(B) Torque, τ = M x B x Sin 90
τ = 0.05 x 0.037 x 1
τ = 1.85 x 10^-3 Nm
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:
yes independent of the sign or valve of Q
Explanation:
Answer:
a) 
b) 
c) Compressing is easier
Explanation:
Given:
Expression of force:

where:



when the spring is stretched
when the spring is compressed
hence,

a)
From the work energy equivalence the work done is equal to the spring potential energy:
here the spring is stretched so, 
Now,
The spring constant at this instant:



Now work done:



b)
When compressing the spring by 0.05 m
we have, 
<u>The spring constant at this instant:</u>



Now work done:



c)
Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.