The final temperature of the gas given the data from the question is 1527 °C
<h3>Data obtained from the question</h3>
- Initial volume (V₁) = V
- Initial temperature (T₁) = 27 °C = 27 + 273 = 300 K
- Final volume (V₂) = 6V
- Final temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The final temperature of the gas can be obtained by using the Charles' law equation as illustrated below:
V₁ / T₁ = V₂ / T₂
V / 300 = 6V / T₂
Cross multiply
V × T₂ = 300 × 6V
Divide both side by V
T₂ = (300 × 6V) / V
T₂ = 1800 K
Subtract 273 from 1800 K to express in degree celsius
T₂ = 1800 – 273
T₂ = 1527 °C
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Answer:
<em>The initial speed of the sprinter was 2.2 m/s</em>
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
The following relation applies:

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.
The sprinter speeds up from an unknown initial speed to vf=3 m/s in t=2 seconds with an acceleration of
.
To find the initial speed, we solve the equation for vo:

Substituting the values:



The initial speed of the sprinter was 2.2 m/s
Answer:
(a) θ1 = 942.5rad, (b) θ2 = 13195 rad
Explanation:
(a) Given
ωo = 0 rad/s
ω = 3600rev/min = 3600×2(pi)/60 rad/s
ω = 377rad/s
t1 = 5s
θ1 = (ω + ωo)t/2
θ1 = (377 +0)×5/2
θ1 = 942.5 rads
(b) ωo = 377rad/s
ω = 0 rad/s
t2 = 70s
θ2 = (ω + ωo)t/2
θ2 = (377 +0)×70/2
θ2 = 13195 rad
Answer:
C ) 1.53
Explanation:
The critical angle of a material is given by the formula

where
c is the critical angle
n is the refractive index
This formula is valid if the second medium is air (which is the case of the problem).
In this problem, we know the critical angle:

Therefore we can rearrange the equation to find the refractive index:
