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PilotLPTM [1.2K]
3 years ago
5

What is wave polarization?

Physics
1 answer:
Anettt [7]3 years ago
7 0
An expression of the orientation of the lines of electric flux in an electromagnetic field 
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A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh
Artyom0805 [142]

Answer:

The electric field is 5.623\times10^{4}\ N/C

Explanation:

Given that,

Radius = 0.357 m

Charge Q=5.03\times10^{-6}\ C

Point charge q=4.15\times10^{-6}\ C

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}

Where, q = point charge

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}

E=5.623\times10^{4}\ N/C

Hence, The electric field is 5.623\times10^{4}\ N/C

4 0
3 years ago
Two charges q1 and q2 exert a 90 N electrostatic force onto each other when they are 1 m apart. They are moved further away to a
Dovator [93]

Answer:

?????????????????????????

6 0
3 years ago
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Practicing soccer skills works most muscles groups.<br><br> True Or False
defon

Answer:

True

Explanation:

4 0
3 years ago
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Can you ever absolutely prove that a hypothesis is correct? Explain.
Diano4ka-milaya [45]

Yes you can, with using scientific experiment.

Ask a question -- Do background Research -- Construct a Hypothesis --Test with an Experiment -- Procedure working? -- Yes or no? -- Analyze Data and Draw Conclusions

With an experiment you can discover if its correct or not.

Hope this helps ! <3

8 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
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