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3 years ago

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

Physics

2 answers:

gtnhenbr [62]3 years ago

8 0

Answer:

Objects that fall near Earth’s surface are rarely in free fall.

"Free fall" is the situation where the ONLY force on an object is

the force of gravity, and nothing else.

Objects near Earth's surface are almost always surrounded by air.

If they are falling, then the air is exerting forces on them, and they

are not in "free fall".

pls mark me brainliest

Explanation:

timurjin [86]3 years ago

6 0

Answer:

Air exerts forces that fall near Earth’s surface

Explanation:

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Select the correct locations on the image.

RideAnS [48]

Answer:

Hurricanes hit almost everywhere but I will say the gulf of for your fourth answer.

Explanation:

you might be right on the north pacific but here is another option.

7 0

3 years ago

Read 2 more answers

Two blocks joined by a string have masses of 6 and 9 kg. They rest on a frictionless horizontal surface. A 2nd string, attached

Tom [10]

Answer:

12N

Explanation:

Suppose the string mass is negligible, the total mass of the 2 block system is 6 + 9 = 15 kg

So the acceleration of the system when subjected to 30N force is

a = F / M = 30 / 15 = 2 m/s2

So both blocks would have the same acceleration, however, the force acting on the 6kg block would have a magnitude of

f = am = 2 * 6 = 12N

This is the tension in the string between the blocks

5 0

3 years ago

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

gizmo_the_mogwai [7]

Let's see

Momentum be P

$\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c$

$\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c$

$\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c$

On comaparing

b-3c=1
b-3=1
b=1+3
b=4

and

-a-b=-1
-a-4=-1
-a=-1+4=3
a=-3

So the unit is

DV⁴/F³

5 0

2 years ago

Read 2 more answers

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

Aleonysh [2.5K]

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

3 0

3 years ago

Frequency, period and wavelength<br> 11th grade high school physics

zzz [600]

Answer:0.38

Explanation:

the formula is f = c / λ

so f= 2.5/6.5

and that equals 0.38 46 and so on so i just rounded it

6 0

3 years ago