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Artist 52 [7]
3 years ago
13

When a quantity of monatomic ideal gas expands at a constant pressure of 4.00×104pa, the volume of the gas increases from 2.00×1

0−3m3 to 8.00×10−3m3?
Physics
2 answers:
3241004551 [841]3 years ago
7 0
This can be verified if we know the values of the initial (T1) and final (T2) temperatures. We use the ideal gas equation for this: PV=RT.

P1V1=RT1
(40000 Pa)*(0.002 m^3) = (8.314 m3Pa/molK)(T1)
T1 = 9.62 K


P2V2=RT2
(40000 Pa)*(0.008 m^3) = (8.314 m3Pa/molK)(T2)
T2 = 33.5 K

Thus, this is true if the monoatomic ideal gas is heated from 9.62 K to 33.5 K at constant pressure.
Semmy [17]3 years ago
5 0
Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
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So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.
bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at²   Subtract ViT from both sides

D - ViT = 0.5at²    Divide both sides by 0.5t²

\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }    I wrote this step out a little more to show how your fraction will cancel

\frac{D - ViT}{0.5t^{2} }= a    I like to flip these around so the single variable is on the right

a = \frac{D - ViT}{0.5t^{2} }

7 0
3 years ago
A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?
Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h  

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3  

V = 6.307 x 10^-5 m^3  

By density = m/V  

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

6 0
3 years ago
A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant fo
Alinara [238K]

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

6 0
3 years ago
A hamster in it's ball starts at rest and accelerates to 3ms1 in 6 seconds.
taurus [48]

Answer:9m

Explanation:

Ball starts from rest . Time taken = 6 seconds. Distance travelled by ball. ∴Distance travelled = 9 m

Hope it helps you

Good luck

7 0
3 years ago
What is the average power consumption in watts of an appliance that uses 5.00 KW.h of energy per day?
abruzzese [7]
(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules

Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts

(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules
3 0
3 years ago
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