The tension in the string and the acceleration must be equal for both masses. (See the free body diagrams)
Answer:
2.35 m/s²
Explanation:
Given that
Mass of the smaller crate, m₁ = 21 kg
Mass of the larger crate, m₂ = 90 kg
Tensión of the rope, T = 261 N
We know that the sum of all forces for the two objects with a force of friction F and a tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.
1) no sliding can also mean that:
a₁ = a₂ = a
This makes us merge the two equations written above together as:
m₂a = T - m₁a
If we then solve for a, we would have something like this
a = T / (m₁+m₂)
a = 261 / (21 + 90)
a = 261 / 111
a = 2.35 m/s²
Therefore, the needed acceleration of the small crate is 2.35 m/s²
Answer:
The driving force for (a) heat transfer is temperature difference. (b) electric current is voltage difference. (c) fluid flow is pressure or hydraulic head difference.
Explanation: (a) The driving force for heat transfer is temperature difference. Heat transfer between two mediums is possible only if the two mediums are at different temperature, the higher the temperature, the higher the heat transfer.
(b) The driving force for electric current is voltage difference. Voltage difference is defined as the potential difference in charge between two points in electrical field. For electric current to occur,the voltage must be high.
(c) The driving force for fluid flow is pressure difference or hydraulic head difference. For fluid to move upward,it requires energy.
Answer:
e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1
Explanation:
This is an ejercise in special relativity, where the speed of light is constant.
Let's carefully analyze the approach, we see the two events at the same time.
The closest event time is
c = (x₁-300) / t
t = (x₁-300) / c
The time for the other event is
t = (x₂- 600) / c
since they tell us that we see the events simultaneously, we can equalize
(x₁ -300) / c = (x₂ -600) / c
x₁ = x₂ - 300
We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1
