If I push on the wall with 75 Newtons of force, the wall will push back with _

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward. what maximum height does she

Mumz [18]

Her center of mass will rise 3.7 meters. First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so 8.5 m / 9.8 m/s^2 = 0.867346939 s And the distance a object under constant acceleration travels is d = 0.5 A T^2 Substituting known values, gives d = 0.5 9.8 m/s^2 (0.867346939 s)^2 d = 4.9 m/s^2 * 0.752290712 s^2 d = 3.68622449 m Rounded to 2 significant figures gives 3.7 meters. Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.

8 0

3 years ago

Heat transfer that doesn’t need a medium? A) Radiation B)Convection C)Conduction

Degger [83]

Radiation is a method of heat transfer through electromagnetic waves. This transfer does not require a medium in order for heat to be transferred.

The solution is A.

8 0

3 years ago

Read 2 more answers

If the speed of light in a substance is 2.26 x 10^8 m/s, what is the index of refraction of that substance?

vladimir1956 [14]

Answer:

1.33

Explanation:

speed of light in vacuum, c = 3 x 10^8 m/s

speed of light in medium, v = 2.26 x 10^8 m/s

The refractive index of the medium is given by

μ = speed of light in vacuum / speed of light in medium

μ = (3 x 10^8) / (2.26 x 10^8)

μ = 1.33

4 0

2 years ago

What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s?

Inga [223]

Frequency = velocity/wavelength

Frequency = 10/20

Frequency = 0.5 Hz

3 0

3 years ago

If I push on the wall with 75 Newtons of force, the wall will push back with __