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Naya [18.7K]
3 years ago
9

Why are objects that fall near Earth’s surface rarely in free fall? Gravity does not act on objects near Earth’s surface. Air ex

erts forces on falling objects near Earth’s surface. The objects do not reach terminal velocity. The objects can be pushed upward by gravity.
Physics
2 answers:
o-na [289]3 years ago
6 0

Answer:

its B. i just took the quiz

Explanation:

Nadya [2.5K]3 years ago
4 0

Answer:

B.Air exerts forces on falling object near earth's surface.

Explanation:

Free fall: When no external force like air resistance act on the body  only gravity force act on the body then we say the body falling freely.

Gravity force :It is that force which act on the center of the body and and body fall towards earth.

Force due to gravity is given by

F=mg

Where

m=Mass of the body

g=Acceleration due to gravity

But when the object that fall near earth's surface then molecules of air exert force on the object .Therefor, air resistance force act on the body near earth's surface.

When the air resistance act on the body then the object that fall near earth's surface is not free fall because both force act on the body force due to gravity and air resistance.

Air resistance force act on the object in opposite to direction of gravity.

B.Air exerts forces on falling object near earth's surface.

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Gnoma [55]
1. turn off lights in classrooms when not in use.
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4. Don't charge your cell phones during class.
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5 0
3 years ago
How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do
sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

x = ut +  \frac{1}{2} a {t}^{2}

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

h =  \frac{1}{2} g {t}^{2}  =  \frac{1}{2}  \times 9.81 \times  {8.7}^{2}  = 371.26m

4 0
3 years ago
When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m
SpyIntel [72]

Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

7 0
3 years ago
when you look outside on a cold morning, you might see dewdrops on leaves. Which process causes dewdrops to form?
AlekseyPX
Condensation causes dewdrops to form
8 0
4 years ago
Read 2 more answers
(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasona
antoniya [11.8K]

Answer:

Part a)

V = 7.2 \times 10^{11} Volts

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Explanation:

Part a)

As we know that the potential near the surface of metal sphere is given by the equation

V = \frac{kQ}{R}

here we have

Q = 8 C

R = 10.0 cm

now we have

V = \frac{(9\times 10^9)(8 C)}{0.10}

V = 7.2 \times 10^{11} Volts

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

3 0
3 years ago
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