Answer:
33.68 N
Explanation:
Data
W= 32J
d- 0.95m
F= ?
W=Fd
They are asking for the magnitude which is the force, so you need to solve for force.
F=W/d
= 32J/ 0.95m
= 33.68 N
Answer:
Explanation:
a )
one kg of coal gives energy of 27 x 10⁶ J
75 kg of coal gives energy of 27 x 10⁶ x 75 J
So rate which energy is coming out of coal per second
= 27 x 10⁶ x 75 J
= 2025 x 10⁶ J /s
2025 million watts .
b ) energy output = 800 million watts
efficiency = (800 / 2025) x 100
= 39.5 % .
Transverse wave as the wave is going up and down no compressions
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
Answer:
Explanation:
Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .
Then length of the platform = length of the train rocket .
A )
Time to cross a particular point on the platform
= length of rocket train / .96 x 3 x 10⁸
= 90 / .96 x 3 x 10⁸
= 31.25 x 10⁻⁸ s
B) Rest length of the rocket = length of platform = 90 m
C ) length of platform as viewed by moving observer =

= 
= 321 m
D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s
for the observer in the rocket , time will be dilated so time recorded by observer in motion ,
8.75 x 10⁻⁸ s .