A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma
2 answers:
The gravitational force on the woman is A) 500 N
Explanation:
There are two forces acting on the woman during her fall:
- The force of gravity,
, acting downward
- The air resistance,
, acting upward
According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:
where m is the mass of the woman and a her acceleration.
The net force can be written as
Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:
Combining the equations together, we get:
which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:
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Answer:
The resultant velocity is 86.1 mi/h.
Explanation:
The law of cosines is given by:
Where:
c: is the resultant velocity =?
a: is the velocity of the plane = 75.0 mi/h
b: is the velocity of the wind = 15.0 mi/h
θ: is the angle between "a" and "b"
The angle between "a" and "b" can be found as follows:
Now, by using the law of cosines we have:
Therefore, the resultant velocity is 86.1 mi/h.
The law of sines is:
Where:
γ: is the angle between "b" and "c"
α: is the angle between "a" and "c"
So, if we want to find "c" by using the law of sines, we need to know another angle besides θ (γ or α), and the statement does not give us.
I hope it helps you!
The rate at which the height is changing is ( 5 / x ) m / hr
We know that,
Area of an equilateral triangle A =
/ 4
h = x / 2
Where,
x = Side
h = Height
Given that,
dA / dt = 5 / hr
h = x / 2
Differentiate both sides with respect to t
dh / dt = ( / 2 ) ( dx / dt )
dx / dt = ( 2 / ) ( dh / dt )
A =
/ 4
Differentiate both sides with respect to t
dA / dt = ( / 4 ) ( 2x ) ( dx / dt )
5 = ( / 4 ) ( 2x ) ( 2 /
) ( dh / dt )
dh / dt = ( 5 / x ) m / hr
Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt
Therefore, the rate at which the height is changing is ( 5 / x ) m / hr
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