Answer:
238.75⁰C .
Explanation:
coefficient of linear thermal expansion of aluminum and steel is 23 x 10⁻⁶ K⁻¹ and 12 x 10⁻⁶ K⁻¹ respectively .
Rise in temperature be Δ t .
Formula for linear expansion due to heat is as follows
l = l₀ ( 1 + α x Δt )
l is expanded length , l₀ is initial length , α is coefficient of linear expansion and Δt is increase in temperature .
For aluminum
l = 2.5 ( 1 + 23 x 10⁻⁶ Δt )
For steel
l = 2.506 ( 1 + 12 x 10⁻⁶ Δt )
Given ,
2.5 ( 1 + 23 x 10⁻⁶ Δt ) = 2.506 ( 1 + 12 x 10⁻⁶ Δt )
1 + 23 x 10⁻⁶ Δt = 1.0024 ( 1 + 12 x 10⁻⁶ Δt )
1 + 23 x 10⁻⁶ Δt = 1.0024 + 12.0288 x 10⁻⁶ Δt
10.9712 x 10⁻⁶ Δt = .0024
Δt = 218.75
Initial temperature = 20⁰C
final temperature = 218.75 + 20 = 238.75⁰C .
Answer:
shadow length 7.67 cm
Explanation:
given data:
refractive index of water is 1.33
by snell's law we have


solving for



from shadow- stick traingle


s = 19tan22 = 7.67 cm
s = shadow length
The prefix of a unit could give you a hint on its extent with respect to the base unit. In this case, the base unit is money in terms of dollars. When you say kilodollars, that would be 1,000 times as large as a dollar. Hence,
33,200 dollars * 1 kilodollar/ 1,000 dollars = 33.2 kilodollars
When you say megadollars, that would be a million times as large as a dollar. Hence,
33,200 dollars * 1 megadollar/1,000,000 dollars = 0.0332 megadollars
Answer:

Explanation:
At the bottom the tension would be upwards and the weight downwards, their difference being the centripetal force. Taking the upwards direction as positive we then have:

where we have used the equation for centripetal acceleration. Thus we have:

Answer:
f = v / 4L
the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
Explanation:
In wind instruments the wave speed must meet
v = λ f
λ = v / f
from v is the speed of sound that depends on the temperature
v = v₀
where I saw the speed of sound at 0ºC v₀ = 331 m/s the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation
(F -32) 5/9 = C
76ºF = 24.4ºC
45ºF = 7.2ºC
With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside
at T₁ = 24ºC v₁ = 342.9 m / s
at T₂ = 7ºC v₂ = 339.7 m / s
To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change
λ / 4 = L
λ= 4L
v / f = 4L
f = v / 4L
Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.