Ask question

Ask question

All categories

katovenus [111]

2 years ago

9

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

evation where the pressure is 0.77 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?

1 answer:

Irina-Kira [14]2 years ago

4 0

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

You might be interested in

A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t

Gelneren [198K]

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

$V=\dfrac{4}{3}\pi r^3$

$V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3$

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb + N = m g

m=Mass of the ball = Density x volume

m = γ V , γ =Density of the Ball

ρ V g + N = γ V g ( take g= 10 m/s²)

$\gamma =\dfrac{N+\rho V g}{V g}$

$\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times 10}{219.13\times 10}$

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

5 0

2 years ago

Which of the following is a scalar quantity

Kay [80]

Answer :

I think it is A. Distance

7 0

2 years ago

HELLO I NEED YOU HELP WITH THIS SCIENCE QUESTION NO LINKS!!!

sweet [91]

the study of how matter and energy interact

the study of the natural world around us

the study of human bodies and how they move

8 0

1 year ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25

torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{V}{d}$

Put the value into the formula

$E=\dfrac{25.0}{1.50\times10^{-3}}$

$E=16666.66\ V/m$

We need to calculate the charge density

Using formula of charge density

$\sigma=E\times\epsilon_{0}$

Put the value into the formula

$\sigma=16666.66\times8.85\times10^{-12}$

$\sigma=1.474\times10^{-7}\ C/m^2$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}$

$C=4.484\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=4.484\times10^{-12}\times25.0$

$q=112.1\times10^{-12}\ C$

Hence, The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

8 0

2 years ago

I need help with these questions ☺️☺️

Mariana [72]

Answer:

1. well the value may vary because of different reactions to the technology because its new to a lot of people but to younger ones its something normal and something we cant live without but with older ones its something that they have lived without for most of there life.

Explanation:

6 0

2 years ago