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SVETLANKA909090 [29]
3 years ago
6

A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2

.7 atm, and the temperature of the gas increases to 298 K. What is the final volume of the gas, rounded to the nearest tenth? 2.1 L 2.5 L 2.7 L 3.0 L
Physics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

5.25/275= 2.7*V2/298

Cross multiply

275*2.7V2= 298*5.25

742.5V2= 1564.5

V2= 1564.5/742.5

V2= 2.10L

Hence the final volume is 2.10L

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Diffrence between:- movable pulley and fixed pully​
kotykmax [81]

Answer:

fixed pulley: A pulley system in which the pulley is attached to a fixed point and the rope is attached to the object. ... movable pulley: A pulley system in which the pulley is attached to the object; one end of the rope is attached to a fixed point and the other end of the rope is free.

Explanation:

8 0
2 years ago
Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia
USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

8 0
3 years ago
Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his
just olya [345]
The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.
4 0
3 years ago
Read 2 more answers
A person travels by car from one city to another with different constant speeds between pair of cities. She drives for 36 min at
Softa [21]

 Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:

let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance 

3 0
3 years ago
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p
Semenov [28]

Answer:

f_n=3.75N

Explanation:

From the question we are told that:

Frictional force F=0.150N

Coefficient of kinetic friction \mu=0.04

Generally the equation for Normal for is mathematically given by

 f_n=\frac{F}{\mu}

Therefore

 f_n=\frac{0.150}{0.04}

 f_n=3.75N

5 0
3 years ago
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