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3 years ago

I hope everyone is ok rn with this tornado . prayers if anyone hurt

Physics

2 answers:

skad [1K]3 years ago

4 0

<h2><em>THERE WAS A TORNADO????????????</em></h2><h2><em></em></h2><h2><em>SAY WHAT?!</em></h2>

Marina86 [1]3 years ago

3 0

I hope everyone that everyone that is in the tornado stays safe. My prayers go out to them! have a nice day.

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Velocities of two bodies A and B are given in vectors notation as va =i+2j-3k and Vb=3i+2j-k what will be the relative velocity

ArbitrLikvidat [17]

Answer:

$V_{B/A}=2i+2k$

Explanation:

The relative velocity can be calculated by means of the difference between vector B minus vector A.

$V_{A}=i+2j-3k\\V_{B}=3i+2j-k\\V_{B}-V_{A}=(3-1)i + (2-2)j+(-1-(-3))k\\V_{B/A}=2i+2k$

5 0

3 years ago

Gravity is a <br><br> A.pushing force<br> B. not a force at all<br> C. pulling force

melamori03 [73]

Answer:

Explanation:

gravity is a pulling force according to Newton

4 0

3 years ago

How do you solve 0.004 dm + 0.12508 dm?

Effectus [21]

0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.

The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.
In this problem, it just happens to be a dm.

7 0

3 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 :-

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

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

5 0

4 years ago

The__ of friction is a number that represents the resistance to sliding

Tresset [83]

Answer:

B. coefficient

Explanation:

i dont have to explain right?

3 0

3 years ago