Answer:
Explanation:
Due to heat energy , metal expands . Formula for linear expansion is as follows .
L = l ( 1 + α Δt )
where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .
To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are
20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .
For steel sphere ,
L = 25 ( 1 + 12 x 10⁻⁶ Δt )
For brass ring
L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )
1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt
.004 = 7.9518 x 10⁻⁶ Δt
Δt = 4000 / 7.9518
= 503⁰C.
final temp = 503 + 15 = 518⁰C .
Answer:
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Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
Answer:
ugmd = 1/2 kx²
d = (1/2 kx²) / (ugm)
= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)
= 7.4 m
ugmd = 1/2 mv²
v = √2ugd
= √(2(0.23)(9.81 m/s²)(7.4 m)
= 5.8 m/s
Explanation:
