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2 years ago

8

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

ed taking the young's modulus to be 2.0×10^11 N/m^2

1 answer:

UkoKoshka [18]2 years ago

7 0

Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

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For 10 minutes, the wolf blew on the brick house with 5,000 N of force. How much work did he do?

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A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged ro

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The sphere is positively charged

Explanation:

This is because when the positively charged rod is brought near the metal rod A, the electrons in metal rod A and sphere B are attracted towards it into metal rod A while the positive charges in the are repelled into sphere B. So, when the charged rod is withdrawn, and metal rod A and sphere B are separated, metal rod A is now negatively charged, but sphere B is positively charged.

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If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be

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3 years ago

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Please help me with this for science

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2 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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3 years ago