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ollegr [7]
3 years ago
5

I need help with the circled one

Physics
1 answer:
Whitepunk [10]3 years ago
7 0
Yes the answer is yes 
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The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre
IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0
3 years ago
Describe a perfect day
kvv77 [185]

Answer:

A bright and sunny day not worrying about work or school no family drama just a day you can relax and be yourself surrounded by the people you love.

hope this helps

have a good day :)

Explanation:

7 0
3 years ago
Read 2 more answers
A helium-filled balloon is launched when the temperature at ground level is 27.8°c and the barometer reads 752 mmhg. if the ball
ololo11 [35]
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.

Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³

Given:
At ground level,
p₁ = 752 mm Hg
     = (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
     = 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
     = 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
     = 300.8 K

At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
     = 9.7326 x 10³ Pa
T₂ = 235 K

If the volume at  36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
     = (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
     = 762.15 m³

Answer: 762.2 m³  
3 0
3 years ago
A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s^2 . She is driving at 20 m/s when su
QveST [7]

Answer:

The car stops after 32.58 m.

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 20 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6 m/s²

Time taken by the car to stop

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s

Total Time taken by the car to stop is 0.5+3.33 = 3.83 s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m

The car stops after 32.58 m.

Distance between car and obstacle is 50-32.58 = 17.42 m

4 0
3 years ago
What is 64 nanometers to m?
defon
\sf Hello!

\sf We\: know \:that,
\sf 1\: meter = \sf 10^{9} nm

\sf Then,

\sf Distance\: in \:m = \sf Distance\: in\: nm × \dfrac{\sf 1}{\sf 10^{9}}\: \sf m

⇒ \sf Distance\: in \:m = \sf 64 × 10^{-9} \:m

⇒ \sf Distance\: in\: m = \sf 6.4 × 10^{-8} \:m

~ \sf iCarl
3 0
3 years ago
Read 2 more answers
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