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3 years ago

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Difference between freefall and weightlessness.

1 answer:

Margaret [11]3 years ago

6 0

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>

When a body falls only under the influence of gravity, it is called free fall.
Freefall is not possible in absence of gravity.
A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>

Weightlessness is a condition at which the apparent weight of body becomes zero.
Weightlessness is possible in absence of gravity.
A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

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alisha [4.7K]

Answer: barometer.

A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.

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3 years ago

What electrical force dies a Uranium nucleus exert on one of its inner electrons, located at a distance of 175 picometers (= 1.7

AlekseyPX

Answer:

correct option is d) 7.0 x 10^-7 N

Explanation:

given data

distance = 175 picometers = 1.75 × $10^{-10}$ m

to find out

electrical force

solution

we know atomic no of uranium is 92

and charge on electron is = 1.6 × $10^{-19}$ C

and electrical force is express as

electrical force = $\frac{1}{4 \pi \epsilon _o} \frac{q1q2}{r^2}$ .............1

put here value we get

electrical force = $9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}$

electrical force = 6.921 × $10^{-7}$ N

so correct option is d) 7.0 x 10^-7 N

5 0

3 years ago

Read 2 more answers

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

vfiekz [6]

The potential difference comes out to be

$10 \times 10 {}^{ - 3} m$

Given:

σ = 8. 85 × 10-9 c/m2

we know,

$E = \frac{σ}{2ε0}$

$E = \frac{8.85 \times 10 {}^{ - 9} }{2ε0}$

$E = \frac{v}{d}$

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

$= \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }$

$Δ = 10 \times 10 {}^{ - 3} m$

Thus the potential difference is

$10 \times 10 {}^{ - 3} m$

Learn more about potential difference from here: brainly.com/question/28165869

#SPJ4

5 0

2 years ago

1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

6 0

3 years ago

What is the purpose of the author’s description of Montresor’s encounter with Fortunato above? The author is giving the reader a

evablogger [386]

Well if the encounter is good then that would be great to write about!!

4 0

3 years ago

Read 2 more answers