Answer:
2m₁m₃g / (m₁ + m₂ + m₃)
Explanation:
I assume the figure is the one included in my answer.
Draw a free body diagram for each mass.
m₁ has a force T₁ up and m₁g down.
m₂ has a force T₁ up, T₂ down, and m₂g down.
m₃ has a force T₂ up and m₃g down.
Assume that m₁ accelerates up and m₂ and m₃ accelerate down.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₁ − T₂ − m₂g = m₂(-a)
T₁ − T₂ − m₂g = -m₂a
(m₁g + m₁a) − T₂ − m₂g = -m₂a
m₁g + m₁a + m₂a − m₂g = T₂
(m₁ − m₂)g + (m₁ + m₂)a = T₂
Sum of the forces on m₃:
∑F = ma
T₂ − m₃g = m₃(-a)
T₂ − m₃g = -m₃a
a = g − (T₂ / m₃)
Substitute:
(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂
(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂
(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂
m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂
2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂
T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)
Vf = final velocity
vo = initial velocity
a = acceleration
t = time
use the following equation
vf = vo + at
since vo = 0 m/s (stopped), that term drops out and you're left with . . .
vf = at
(60 m/s) = (8.0 m/s²)t
t = (60 m/s)/(8.0 m/s²) = 7.5 seconds
<u><em>t = 7.5 seconds</em></u>
Hey!
Given ,
Acceleration, a = 5 m/s^2
time , t = 10 seconds
Initial velocity,u = 0
Final velocity , v = ?
We have ,
v = u + at
=> v = (0)+(5)(10)
=> v = 50 m/s
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity
d= 250s
t= 17s
a= d/t
=
= 4.7
The bullet would start to veer of path, either to the right or left and also downword. This happens because gravity is pushing it down and it no longer has enough thrust force pushing it in a straight direction because drag and air friction are slowing it down. The forces would include gravity, thrust, drag, and air resistence.