A 45000 watt crane operating at full power lifts a 2100 kg object vertically for 17.4 seconds. How high has the crane lifted the
1 answer:
Answer:
Explanation:
We need the power equation here, which is:
Power = (F * Δx)/time
where F * Δx is the amount of work done.
F is a force which is measured in Newtons. We are given the mass of the object, but since we need a Force measure, we need the weight of the object:
F = 2100(9.0)
F = 21000 to the correct number of sig dig.
Now we can plug in the values we have and solve for the displacement, Δx:
and isolating x:
so
x = 37 m
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Answer: a) 11.76 m/s b) 7.056 m
Explanation:
The described situation is as follows:
An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.
This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:
(1)
(2)
Where:
Is the final velocity of the object
Is the initial velocity of the object (it was dropped)
is the acceleration due gravity
is the height of the tower
is the time it takes to the object to reach the ground
b) Begining with (1):
(3)
(4)
(5) This is the final velocity of the object
a) Substituting (5) in (2):
(6)
Clearing :
(7)
(8) This is the height of the tower
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N