Work = (force) x (distance)
The work he did: Work = (700 N) x (4m) = 2,800 joules
The rate at which
he did it (power): Work/time = 2,800 joules/2 sec
= 1,400 joules/sec
= 1,400 watts
= 1.877... horsepower (rounded)
Answer:
B. Marginal cost equals long-run average total cost.
Explanation:
The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.
Answer:
Any Lens
Explanation:
I Hope it's right if not so Sorry :)
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
Now, we can calculate the final speed of the crate at the end of 10.0 m:
For the next 10.5 meters we have frictional force:
So, the acceleration is:
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!
