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MrRissso [65]
3 years ago
11

Why are objects that fall near Earth's surface rarely in free fall?

Physics
1 answer:
sveta [45]3 years ago
6 0

Answer: Air exerts forces on falling objects near Earth's surface.

Explanation: Objects falling near the earth surface are rarely in free fall due to the force exerted by air on falling object near the earth( air resistance).

The acceleration of free fall tend to pull the body towards the earth surface while air resistance (drag) tends to act in the opposite direction.

Given that the weight of the body is always  constant.

The drag acts in the upward direction, thereby negating the downward weight of the object.

From Newton's Second law:

Force = mass × acceleration due to gravity

Acceleration = force / mass

The net force acting on the object becomes :

(Downward weight - upward drag)

This hampers the free fall of the object due to gravity.

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Two different wave groups with the same height travel together in the same direction. The wavelength of one group is twice as lo
Morgarella [4.7K]

Answer:

A combined wave of extra height will produce every other wave.

Explanation:

One wave Crest can catch up to another wave Crest if the group of waves are from different places.

Here the two group of waves have different wavelength but travelling together in the same direction so when they combine,they will produce every other wave.

8 0
4 years ago
How is energy conserved in a process of a heart beat
Westkost [7]

Explanation:

energy conservation and fatigue management -tiredness is a common symptom of a heart attack and although rest is important activity is also required to facilitate a return to health. an occupational therapist said energy conservation and fatigue management is techniques to be implemented throughout the day. to help clients achieve their goals

3 0
4 years ago
A 4.00 µf capacitor is connected to a 12.0 v battery.
atroni [7]
(a) The capacitance of the capacitor is:
C=4 \mu F=4 \cdot 10^{-6}F
and the voltage applied across its plates is
V=12.0 V

The relationship between the charge Q on each plate of the capacitor, the capacitance and the voltage is:
C= \frac{Q}{V}
and re-arranging it we find the charge stored in the capacitor:
Q=CV=(4 \cdot 10^{-6} F)(12.0 V)=4.8 \cdot 10^{-5} C

(b) The electrical potential energy stored in a capacitor is given by
U= \frac{1}{2}CV^2
where C is the capacitance and V is the voltage. The new voltage is 
V=1.50 V
so the energy stored in the capacitor is
U= \frac{1}{2}(4 \cdot 10^{-6} F)(1.50 V)^2=4.5 \cdot 10^{-6} J
3 0
3 years ago
Read 2 more answers
A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i
nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

U=mgh

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

U=mgh

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J

3. 0 J

As before, the gravitational potential energy of the ball is given by

U=mgh

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J

4 0
4 years ago
What is the relative energy expense of galloping rather than trotting at 3.5 m/s?
labwork [276]

Answer: Trotting uses only 75 percent of the energy as galloping

Explanation: Trotting is only 300 J/m, whereas galloping is roughly 400 J/m

7 0
3 years ago
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