Let volume of empty boat be = 100% = 1V
and mass of boat be M
In water 10%, 0.1V of the volume is submerged.
Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V
M leads to 0.1V boat submerging
boat submerging.
M + 1200kg leads to 0.7V boat submerging.
This is 60%, 0.6 V increase
By comparison
(M+1200kg) * 0.1V = 0.7V * M
0.1M + 120kg = 0.7M
120kg = 0.7M - 0.1M
120kg = 0.6M
M = (120/0.6)kg
M = 200kg.
The mass of the boat is 200kg.
Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
You should just ask the wave
Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s
D. distance
A light-year is the distance light would travel in 1 year.
