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3 years ago

5

The pressure of a gas changes from 120 kPa. The volume changes from 45 L to 40 L. If the initial temperature is 81 c, what is th

e final temperature in C?

1 answer:

irina [24]3 years ago

7 0

Explanation:

we are not given the pressure change, check yhe question please

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A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capac

Pavlova-9 [17]

Answer:

k = 6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

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2 years ago

A chemical lab isolates a gaseous element from the atmosphere that does not form chemical compounds. In which group of the perio

Sholpan [36]

Answer:

A- 18

Explanation:

It wont let me explain how

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3 years ago

What heat transfer do sun heats earth go through

Eddi Din [679]

Yes because the heat is to hot and would melt the earth layers so it would go through

Brainiest plz

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2 years ago

An object has traveled 20 meters in 40 seconds what is its average speed

Mice21 [21]

Hello,

To solve we need to know the formula for speed

The formula is D/T=S (Distance of time=speed)

Now all we have to do is plug in the numbers.

20/40= 1/2 or 0.5

SO the speed is 0.5 m/s

Have a great day!

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3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago