(a)
1 answer:
a) The momentum of the coconut is 3 kg m/s
b) At first, the air resistance is negligible, so the coconut accelerates due to the force of gravity
c) The coconut reaches its terminal velocity
Explanation:
a)
The momentum of an object is given by the equation
where
m is the mass of the object
v is its velocity
For the coconut in this problem, we have:
m = 1.5 kg (mass)
v = 2 m/s (velocity)
Therefore, its momentum is
B)
There are only two forces acting on the coconut during its fall:
- The force of gravity, of magnitude
(m= mass of the coconut, g = acceleration of gravity), acting downward
- The air resistance, acting upward, whose magnitude is proportional to the speed of the coconut
During the first momentums of the fall, the speed of the coconut is still low, so the air resistance is mostly negligible, and therefore only the force of gravity is acting on the coconut. Since this force is constant, it means that the acceleration of the coconut is constant: therefore, its velocity keeps increasing during the fall, and the coconut speeds up.
C)
If the tree is very tall, the fall of the coconut lasts long, and the speed of the coconut keeps increasing. Since the air resistance is proportional to the speed, this means that at some point, the air resistance is no longer negligible, and it starts to have some effect on the fall of the coconut. In particular, at a certain point, the air resistance will become equal (in magnitude) to the force of gravity (but opposite in direction): this means that from this point, the acceleration of the coconut will be zero, and therefore the coconut will continue its motion at constant velocity. This velocity is called terminal velocity, and it occurs when the force of gravity is equal to the air resistance:
where is the air resistance.
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consider the motion in x-direction
= initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
= acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X = t + (0.5)
t²
100 = (4.60)
= 21.7 m/s
consider the motion along y-direction
= initial velocity in y-direction = ?
Y = vertical displacement = 0 m
= acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y = t + (0.5)
t²
0 = (4.60) + (0.5) (- 9.8) (4.60)²
= 22.54 m/s
initial velocity is given as
= sqrt((
)² + (
)²)
= sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg
ANSWER
EXPLANATION
First, we have to make a sketch of the direction of the moments of the forces about 12m from the left in the diagram:
The sum of upward forces must be equal to the sum of downward forces. This implies that:
Also, the sum of clockwise moments must be equal to the counter-clockwise moments:
From the first equation, make TA the subject of the formula:
Substitute that into the second equation:
Substitute that into the equation for TA:
Therefore, the reaction supports at TA and TB are:
