____________________ (is/are) one type of tectonic event captured in geologic maps.

If you use such a tank to fill 0.020 m3 foil balloons (which don't stretch, and so have an internal pressure that is very close

Arte-miy333 [17]

The question is incomplete, here is a complete question.

Party stores sell small tanks containing 41 g of helium gas.

If you use such a tank to fill 0.020 m³ foil balloons (which don't stretch, and so have an internal pressure that is very close to atmospheric pressure), how many balloons can you expect to fill? Assume the temperature is 20°C.

Answer : The number of balloon filled can be, 12 balloons.

Explanation :

First we have to calculate the number of moles of gas in balloon.

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm (atmospheric pressure)

V = volume of gas = $0.020m^3=20.0L$ $(1m^3=1000L)$

T = temperature of gas = $20^oC=273+20=293K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (20.0L)=n\times (0.0821L.atm/mol.K)\times (293K)$

$n=0.831mol$

Thus, the number of moles of gas in balloon is 0.831 mol per balloon.

Now we have to calculate the mass of gas in balloon.

$\text{Mass of gas}=\text{Moles of gas}\times \text{Molar mass of gas}$

Molar mass of He gas = 4 g/mole

$\text{Mass of gas}=(0.831 mol/balloon)\times 4g/mol=3.32g/balloon$

Now we have to calculate the number of balloon can be filled.

Number of balloon filled = $\frac{41g}{3.32g/balloon}=12.3balloon\approx 12balloons$

Therefore, the number of balloon filled can be, 12 balloons.

8 0

3 years ago

To understand the concept of intensity; the relationship between the power of the source and the intensity of the wave; and the

frez [133]

Answer:

a. 4.77 W/ $m^{2}$

b. 2.122 W/ $m^{2}$

c. 13.78 W/ $m^{2}$

d. 69 km

Explanation:

a.

power P of speaker = 60 W

distance d of speaker = 1.0 m

intensity I of speaker = P/4π $d^{2}$

I = 60/(4 x 3.142 x $1^{2}$ ) = 60/12.568

Intensity = 4.77 W/ $m^{2}$

b.

at a distance of 1.5 m, intensity will be

I = P/4π $d^{2}$

I = 60/(4 x 3.142 x $1.5^{2}$ ) = 60/28.278

Intensity = 2.122 W/ $m^{2}$

c.

combined intensity of the two front speakers will be 2 x 4.77 = 9.54 W/ $m^{2}$

combined intensity of the two back speaker will be 2 x 2.122 = 4.244 W/ $m^{2}$

total sound intensity perceived by the driver will be the superimposition of these four speakers.

I = 9.54 + 4.244 = 13.78 W/ $m^{2}$

d.

Minimum discernible intensity of sound = $10^{-12}$ W/ $m^{2}$

each speaker produces 0.06 W of power.

We assume that each speaker spreads outward evenly.

from I = P/4π $d^{2}$ ,

d = $\sqrt{\frac{P}{4*\pi*I } }$

d = $\sqrt{\frac{0.06} {4*\pi*10^{-12 } }$ = 69094.35 m ≅ 69 km

4 0

3 years ago

What is 1/2 / 1/4. Please help am getting two answers.idk the correct one.

Alla [95]

Answer:

2

hope this helps :)

5 0

3 years ago

A car is traveling at 10 m/s starts to decelerate steadily. It comes to a complete stop in 20 seconds what is it’s acceleration

Ludmilka [50]

Answer:

-0.5 m/s

Explanation:

given:

u (initial velocity) = 10 m/s

v (Final velocity) = 0 m/s (because it is given that it comes to a complete stop)

t (time taken) = 20 s

therefore,

acceleration = v - u / t

a = 0 - 10 / 20

a = -10 / 20

a = -1/2

a = -0.5

There is a acceleration of -0.5 m/s

3 0

3 years ago

Read 2 more answers

Someone pls help mee

Alex17521 [72]

This claim isn’t true. This claim is trying to say that once force is being applied to Newton’s Cradle then it will forever stay in motion. However, from the excerpt we learn that this isn’t possible. As one of the balls are pushed, it is set into kinetic energy, and then that ball will hit another and send it into kinetic energy as well. However, not all of the kinetic energy is kept through this process, some of the energy is lost and converted into different forms such as sound energy. Therefore, it isn’t possible for Newton’s Cradle to stay in motion forever.

3 0

3 years ago