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3 years ago

Why is acceleration negative in a free fall?

2 answers:

6 0

Technically speaking altitude is measured from the ground up.
Therefore, say you're falling from 1000 metres, that number decreases till it reaches 0. Therefore, you're accelerating but towards the ground i.e. towards 0.

Therefore acceleration is negative because you're travelling at -x metres per second and accelerating at -x metres per second^2

murzikaleks [220]3 years ago

5 0

Only because when most people work with vertical motion, they call the upward direction positive. So falling objects have negative velocity and acceleration. But if you prefer to call the downward direction positive, you have every right to do that. Then the force of gravity on a falling object, as well as its velocity and acceleration, are all positive, and its negative altitude steadily grows until it hits the ground.

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In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms t

saveliy_v [14]

Answer:

0.00493 m/s

Explanation:

T = Temperature of the isotope = 85 nK

R = Gas constant = 8.341 J/mol K

M = Molar mass of isotope = 86.91 g/mol

Root Mean Square speed is given by

$v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s$

The Root Mean Square speed is 0.00493 m/s

6 0

4 years ago

When is orange is the new black episodes come out

xenn [34]

Some are out but will be on Netflix June 17

3 0

4 years ago

Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri

mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$ = 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

Here, F = the force ,

dT = time interval over which the force is applied for

= 1.65 s

dP = change in momentum

dP = m x dV

$= m \times [V_f - V_o]$

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

= 218 N

8 0

3 years ago

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

$^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X$

$^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr$

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

$^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X$

$^{211}_{83}Bi\Rightarrow ^{211}_{84}Po$

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

$^{22}_{11}Na\Rightarrow ^{22}_{11-1}X$

$^{22}_{11}Na\Rightarrow ^{22}_{10}Ne$

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

$^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc$

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0

3 years ago

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

<em>where</em>,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

3 0

3 years ago