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)
Given parameters:
Distance to hardware shop = 5 miles
Time to reach hardware shop = 10 minutes
Time spent at the shop = 30 minutes
Average speed to customer home = 45 mph
Time taken for the travel = 20 minutes
Unknown:
Average speed of the contractor = ?
Solution:
Average speed is the total distance covered divided by the total time taken.
Average speed = 
total distance = distance to hardware shop + distance to customer's home
We do not know the distance to customer's home but we have been given the speed and time, so we can find the distance.
Distance = speed x time
Convert the time to hrs and solve;
60 minutes = 1 hr
20 minutes = 
= 
So, Distance = 45mph x = 15miles
Now;
Total distance = 5 + 15 = 20miles
Total time = time to reach hardware shop + time to reach customer's house
= 10 + 20
= 30 minutes
Convert the time from minutes to hrs;
60 minutes = 1hr
30 minutes = 
= 0.5hr
So;
Average speed = 
= 40mph
The average speed is 40mph
Answer:
The tension force in the supporting cables is 7245N
Explanation:
There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:
We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:
The tension force F in the supporting cables is 7245N
Answer:
500000000
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