The temperature to which it must be heated in order to fit the shaft is 73.33 ⁰C.
<h3>
Linear expansivity </h3>
The temperature to which it must be heated in order to fit the shaft is calculated as follows;
where;
- ΔT is change in temperature
- ΔL is change in length = 50.04 mm - 50 mm = 0.04 mm
- α is coefficient of linear expansion
- L is original length
ΔT = (0.04)/(50 x 15 x 10⁻⁶)
ΔT = 53.3 ⁰C
<h3>Final temperature</h3>
T₂ - T₁ = ΔT
T₂ = ΔT + T₁
where;
- T₂ is final temperature
- T₁ is initial temperature
T₂ = 53.3 + 20
T₂ = 73.33 ⁰C
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We do not feel the gravitational forces from objects other than the Earth because they are weak.
3.5 Newton of force is needed to accelerate 140g of ball at 25 m/s^2 using the formula f=ma
Change 140g to kg
Answer:
The 15 ⁰C measured at this altitude is above the standard temperature for the altitude.
Explanation:
The standard temperature at sea level is 15 degrees Celsius. It decreases about 2 degrees C (or 3.5 degrees F) per 1,000 feet of altitude above sea level.
235 meters is equal to 771 feet.
Using the formula below, we can estimate temperature loss due to this change in altitude, that is 771 feet above sea level.
temperature loss = (3.5 x Change in altitude)/1000ft
temperature loss = (3.5 x 771ft)/1000ft = 2.7⁰F, (32 -2.7 = 29.3 ⁰F)
this is equivalent to 1.5⁰C temperature loss.
Thus, the standard temperature of the engineering quadrangle at 235 meters above sea level is 13.5 ⁰C.
Therefore, the 15 ⁰C measured at this altitude is above the standard temperature for the altitude.