Answer:
6227.866 N
Explanation:
F = G . m(goku) . m(planet) / d²
F = 6.674 x 10-¹¹ x 62 x 1.458 . 10¹⁵ / 31²
F = 6227.866 N
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation




Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation


Hence, The maximum height above the point of release is 11.653 m.
The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material
<u>Explanation:</u>
When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.
But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.
Answer: 9.9%
Explanation: efficiency = (work output /work input) × 100
Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.
Work output = 9200 j and work input = 92048 j
Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%
Answer:
9.01amp
Explanation:
Power = V^2/R
Given that v = 11volts, P = 99watts
99 = 11^2/R
11×11 = 99R
121= 99R
R = 121/99
R= 1.22ohms
From ohms Law; V = IR
11volts = I × 1.22ohms
I = 11/1.23
I = 9.01 amp