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)
You said "<span>A rocket's acceleration is 6.0 m/s2.".
That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.
If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.
How long does it take to reach 42 m/s ?
Well, how many times does it have to go 6 m/s FASTER
in order to build up to 42 m/s ?
That's just (42/6) = 7 times.
Writing it correctly, with the units and everything, it looks like this:
(42 m/s) / (6 m/s</span>²)
= (42/6) (m/s) / (m/s²)
= (42/6) (m/s · s²/m)
= 7 seconds
Electricity. I took something like this hope this helps :)
Answer:
the ball didnt hit my face so
Explanation:
Answer:
B
Explanation:
the graph shows the line going up (accelerating) and it isn't curving like d so it doesn't stop accelerating
Hope this helps :)