Answer:
v₂> v₃ velocity canoe is more than velocity fishing boat
Explanation:
For this exercise we must define a system consisting of the girl, Sally and the boat, in one case the canoe and in the other the fishing boat; for this system we can use moment conservation
Initial moment. Before the jump
p₀ = (M + m₂) v
Final moment. After the jump
= M v₁ - m₂ v₂
Where m and v are the masses and speed of the canoe
p₀ = p_{f}
(M + m₂) v = M v₁ - m₂ v₂
In the case of changing the canoe for the heaviest fishing boat, the final moment is
p_{f} = M v₁ - m₃ v₃
p₀ = p_{f}
(M + m₃) v = M v₁ - m₃ v₃
Since the canoe is stopped the speed v = 0, we write the speed of each boat
Canoe
0 = M v₁ - m₂ v₂
v₂ = M / m₂ v₁
Fishing boat
0 = M v₁ - m₃ v₃
v₃ = M / m₃ v₁
Since the masses of the fishing boat (m₃) is greater than the mass of the canoe (m₂) the speed of the fishing boat is less than the speed of the canoe, we can find the relationship between the two speeds
v₂ / v₃ = m₃ / m₂
Here you can see what v₂> v₃ velocity canoe is more than velocity fishing boat
Answer:
Explanation:
energy stored in spring initially
= kinetic + potential energy of block + energy dissipated by friction
= 1/2 mv² + mgh + μ mgcosθ x d
m is mass , v is velocity at top position , h is vertical height , μ is coefficient of friction ,θ is angle of inclination of plane
= m (1/2 v² + gh + μ gcosθ x d )
= 1.05 ( .5 x 5.1² + 9.8 x 4.9 sin35 + .55 x 9.8 cos35 x 4.9 )
= 1.05 ( 13.005 + 27.543 + 21.635)
= 65.3 J .
Answer:
2 hrs 45 minutes
Explanation:
You have to minus one hr because of tome change
This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:
So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).
The kinetic energy of a mass m moving with a velocity v is given by:
And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:
From this we can determine the speed of the mass:
The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).
