Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Answer:
12.7m/s
Explanation:
Given parameters:
Mass of diver = 77kg
Height of jump = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we apply the motion equation below:
v² = u² + 2gH
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
H is the height
Now insert the parameters and solve;
v² = 0² + 2 x 9.8 x 8.18
v = 12.7m/s
Answer:
you are making an observation that uses numbers.
Explanation:
Answer:
1200 W
Explanation:
Power is given by the ratio between work done and time taken:
where W is the work done and t the time taken.
In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is
