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UNO [17]
3 years ago
11

A weather balloon is inflated to a volume of 27.6 l at a pressure of 736 mmhg and a temperature of 26.1 âc. the balloon rises in

the atmosphere to an altitude where the pressure is 360. mmhg and the temperature is -14.0 âc.
Physics
1 answer:
Arisa [49]3 years ago
8 0
We know from gas equation: 

PV=mRT

Where:
P is pressure
V is the volume of the balloon
m is the mass of gas in the balloon (constant)
R is universal gas constant divided by mean molar wt of air (about 28 g/mol) T is thermodynamic temperature (T in Kelvin; T=273 + t (in deg C)
P1 * V1 =m*R*T1 ---- (i) P2 * V2 =m*R*T2 ---- (ii)
Dividing 1 by 2 we get: 
V1 / V2 = (P2/P1) * (T1/T2)
Given:V1=27.6, P1=736mmhg, T1 = 273+26.1=299.1V2=?, P2 = 360mmhg, T2 = 273+(-14)=259
so, V2 = 27.6*(736/360)*(259/299.1) = 48.86
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AnnyKZ [126]

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

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In which state of matter are water molecules measured as having a comparatively high temperature?
AnnZ [28]
Liquid water because if it said very high then it would be water vapor but it didn’t say that so the answer is B liquid water
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Darya [45]

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Explanation:

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A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant
GREYUIT [131]

Your diagram should include four forces:

• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)

• the normal force, pointing up (mag. <em>n</em>)

• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")

• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

<em>n</em> + (-<em>w</em>) = 0

and

<em>p</em> + (-<em>f</em> ) = 0

So then the forces have magnitudes

<em>w</em> = 43.2 N

<em>n</em> = <em>w</em> = 43.2 N

<em>p</em> = 6.30 N

<em>f</em> = <em>p</em> = 6.30 N

5 0
3 years ago
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

v_{rms,o2}^2 *M_{o2} =3RT_{o2}

solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}

T_{o2} =1.58*10^5 K

7 0
3 years ago
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