The final speed of the orange is 7.35 m/s
Explanation:
The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration towards the ground. So we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time elapsed
For the orange in this problem, we have
u = 0 (it is dropped from rest)
is the acceleration
Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:
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Answer:
the heart would fail to efficiently pump oxygenated blood to the body and lungs
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
a= 0.5m/s^2
Explanation:
Force applied on an object is known as
F=m.a (Newton's second law states it)
a=F/m
a=5/10=0.5m/s^2
Answer:
B. changing by a constant amount each second
Explanation:
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