All categories

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:

You might be interested in

3. How do you represent the strength of a force in a free-body diagram?

ivann1987 [24]

Generally, the length of the line will indicate how strong the force is. If you have two opposing forces and one is higher than the other, you would draw the line of the higher force visibly longer.

3 0

4 years ago

I’m stuck in question 2. Please help

kicyunya [14]

Answer:

3 parts see pic

Explanation:

see pic

3 0

3 years ago

Simpson drives his car with an average velocity of 85 km/h eastward. How long will it take him to drive 560 km on a perfectly st

Nutka1998 [239]

hahahahahahhha

wag kayo umasa sa brainly

7 0

3 years ago

Read 2 more answers

Monochromatic light is incident on a grating that is 75 mmwide

statuscvo [17]

Answer:

Explanation:

Given

Grating Width $w=75 mm$

no of lines $N=50,000 lines$

inclination of second order maxima $\theta =32.5^{\circ}$

Condition of maxima For Diffraction grating

$d\sin \theta =m\lambda$

where $d=Grating\ constant$

$\lambda =wavelength$

$m=order\ of\ maxima$

$d=\frac{width}{N}$

$d=\frac{75}{50,000}$

$d=0.0015\ mm$

therefore for second order maxima wavelength of light is given by

$\lambda =\frac{d\sin \theta }{m}$

$\lambda =\frac{0.0015\times \sin (32.5)}{2}$

$\lambda =403\times 10^{-6}\ m$

$\lambda =403\ nm$

3 0

3 years ago

An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for

AURORKA [14]

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

$E = Blv$

where $E$ is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

$l$ is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

$v$ is the velocity of the fluid through the field = ?

$B$ is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x $v$

2.88 x 10^-3 = 1.44 x 10^-3 x $v$

$v$ = 2.88/1.44 = 2 m/s

8 0

4 years ago