You are given a 5kg ball and an 8 kg box. which is true?

The Milky Way galaxy (pictured above) is an example of a(n) __________________ galaxy. At the center of the galaxy is a galactic

OlgaM077 [116]

I think the answer is A.

8 0

3 years ago

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Answer both and i’ll venmo u rn if ur right pls pls

Masteriza [31]

Answer:

2.5

Explanation:

25-22.5n= will give you 2.5

5 0

3 years ago

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A 1500-nF capacitor with circular parallel plates 2.0 cm in diameter is accumulating charge at the rate of 40.0 mC/s at some ins

Delvig [45]

Answer:

$B= 1.14 X 10^{-7} Tesla$

Explanation:

B at point 7.0 cm radially outward is to find out?

Now for field strength in case of parallel plates in given as

$B=\frac{I\mu}{2r\pi}$

$B=\frac{4\pi X 10^{-7} \frac{Tm}{A}X40 X10^{-3} \frac{C}{s}}{2X0.07mX\pi}$

$B= 1.14 X 10^{-7} Tesla$

For the second part as Capacitor is fully charged no more flow of current could take place hence

I= Current =o

so B = 0

6 0

3 years ago

Noise pollution can lead to the disruption of speech communication in children<br> true or false

LenaWriter [7]

False, because I doesn’t matter if there is noise pollution the child will still able to learn the way words work

7 0

3 years ago

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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