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True [87]
3 years ago
5

What happens to the energy of gas particles when an elastic collision takes place?

Physics
2 answers:
seraphim [82]3 years ago
8 0
The third option because the person above me said so
Gnesinka [82]3 years ago
7 0

Answer:

answer 3

Explanation:

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Compare and contrast what happens to an object's motion when balanced or
mario62 [17]

Answer: The motions shifts.

5 0
2 years ago
An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De
gayaneshka [121]
We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
F=T-W
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
a= \frac{F}{m}
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
F=T-W=6000 N-5000 N=1000 N
To find m, we can use the weight of the system. In fact, the weight of an object is given by
W=mg
where g=9.81 m/s^2. Solving for m, and using W=5000 N, we find
m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg

and at this point, we can calculate the acceleration of the system (elevator+people):
a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2
and the acceleration has the same direction of the resultant force, so upward.
8 0
3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
A 10 m long clothesline is strung so that it is perfectly horizontal. when a shirt is hung in the exact center the line sags to
Leto [7]

Answer:

x = 0.0873 m

Explanation:

given,

length of clothesline = 10 m

the line sags to create an angle that is 1 degree below the horizontal on each end.

As mass is hang at the center the angle made let the deflection be 'x'

as the shirt is hang at the center distance will be equal to 5 m.

now

tan \theta = \dfrac{d_e}{d}

tan \theta = \dfrac{x}{5}

tan 1^0 = \dfrac{W}{5}

W = 5 \times 0.0175

x = 0.0873 m

Hence, the cloth line will be x = 0.0873 m from the original length of the clothesline.

3 0
3 years ago
A star exhibits a parallax shift of .30” rounded to 2 significant digits. That star is ____ light years away.
Lubov Fominskaja [6]

AnsweA 2.25 light years away

Explanation:

8 0
3 years ago
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