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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
Learn more about linear expansivity here: brainly.com/question/14325928
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Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x =
Similarly velocity along y direction V_y(1) =
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²
Answer:
v = 7.69 x 10³ m/s = 7690 m/s
T = 5500 s = 91.67 min = 1.53 h
Explanation:
In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:
where,
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Me = Mass of Earth = 5.97 x 10²⁴ kg
r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m
v = orbital speed = ?
Therefore,
<u>v = 7.69 x 10³ m/s</u>
For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.
So, its orbital speed can be given as:
where,
T = Time Period of Satellite = ?
Therefore,
<u>T = 5500 s = 91.67 min = 1.53 h</u>