Answer:
(a) 490 N on earth
(b) 80 N on earth
(c) 45.4545 kg on earth
(d) 270.27 kg on moon
Explanation:
We have given 1 kg = 9.8 N = 2.2 lbs on earth
And 1 kg = 1.6 N = 0.37 lbs on moon
(a) We have given mass of the person m = 50 kg
As it is given that 1 kg = 9.8 N
So 50 kg = 50×9.8 =490 N
(b) Mass of the person on moon = 50 kg
As it is given that on moon 1 kg = 1.6 N
So 50 kg = 50×1.6 = 80 N
(c) We have given that weight of the person on the earth = 100 lbs
As it is given that 1 kg = 2.2 lbs on earth
So 100 lbs = 45.4545 kg
(d) We have given weight of the person on moon = 100 lbs
As it is given that 1 kg = 0.37 lbs
So 100 lbs 
Answer:
It takes millions sometimes hundreds of millions Explanation:
Answer:
Explanation:
Energy stored in a capacitor
= 1/2 CV²
C is capacitance and V is potential of the capacitor .
When capacitor is charged to 24 V ,
E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J
When it is charged to 12 volt
E₂ = 1/2 CV²
.5 X 2.4 X 12 X12
= 172.8 J
Answer: elastic potential energy = 20.27 J
Explanation:
Given that the
Mass M = 0.470 kg
Height h = 4.40 m
Spring constant K = 85 N/m
The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.
But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.
That is
K .E = P.E = mgh
Where g = 9.8m/s^2
Substitutes all the parameters into the formula
K.E = 0.470 × 9.8 × 4.4
K.E = 20.27 J
Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.
Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
