Ask question

Ask question

All categories

3 years ago

14

A fixed mass of an ideal gas has a volume of 800 cm3 under certain conditions. The pressure (in kPa) and temperature (in K) are

both doubled. What is the volume of the gas after these changes with other conditions remaining the same

1 answer:

Mashutka [201]3 years ago

3 0

Answer:

800cm3

Explanation:

This is a general gas law question

Which has the relationship P1V1/T1 = P2V2/T2

P1= P1 in kPa

T1 = T1 in K

V1= 800cm3

V2=?

P2= 2* P1

T2= 2*T1

The volume of gas after the changes (V2), making it as the subject of formula

V2= P1*V1*T2/P2*T1

V2 = P1 *800* 2T1 / 2P1 * T1 ; dividing accordingly, we have

V2 = 800cm3

You might be interested in

Which is the symbol for gold?

OLga [1]

Answer:

Au

Explanation:

Look at a periodic table it can help you a lot.

6 0

3 years ago

Read 2 more answers

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

emmainna [20.7K]

Answer:

$5,2$

Explanation:

From the question we are told that:

Speed of light $C=3.0×10^8 m/s.$

Generally the equation for Average Speed is mathematically given by

$V_{avg}=\frac{d}{t}$

Where

d=Distance between the Earth and the sun

$d=1.5*10^11m$

Therefore

$t=\frac{d}{V_{avg}}$

$t=\frac{1.5*10^11m}{3.0×10^8 m/s.}$

$t=5*10^2s$

Since m and n is given in the form of

$m*10^n$

Therefore

$m=5 & n=2$

$5,2$

3 0

3 years ago

1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

1.

109.6 cm , - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

3 years ago

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0

2 years ago

A car with a mass of 2200 kg is travelling at a rate of 55 m's. What is the cars momentum? *

Varvara68 [4.7K]

Mass=2200kg
Velocity=55m/s

$\\ \sf\bull\dashrightarrow Momentum=Mass\times Velocity$

$\\ \sf\bull\dashrightarrow Momentum=2200(55)$

$\\ \sf\bull\dashrightarrow Momentum=121000kgm/s$

7 0

2 years ago

Read 2 more answers