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2 years ago

Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?

Physics

1 answer:

oksano4ka [1.4K]2 years ago

6 0

Answer:

Your project goes well.

Explanation:

Because that's how it works.

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A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,

stealth61 [152]

The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it. That's sad and
disappointing, but I think the question can be answered without seeing
the picture.

The net force on the crate is zero. Evidence for this is that fact that
the crate is just sitting there. If the net force on an object is not zero,
then the object is accelerating ... it's either speeding up, slowing down,
or its the direction of its motion is changing. If none of these things is
happening, then the net force on the object must be zero.

4 0

2 years ago

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 :-

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

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

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3 years ago

What is a load force

vladimir2022 [97]

Answer:

A push or pull exerted on an object

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3 years ago

Why is Pluto no longer a planet?

Step2247 [10]

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2 years ago

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Where does energy come from that fuels our physical activity

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3 years ago

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