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4 years ago

A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.

Physics

1 answer:

Natali5045456 [20]4 years ago

7 0

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

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3 years ago

A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At

STALIN [3.7K]

Answer:

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Flow Speed at the inlet pipe will be given by using Continuity Equation

$Q_{1}=Q_{2}$

$v_{1}A_{1}=v_{2}A_{2}$

$v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})$

Where, A = area of pipe

$A=\pi\times \dfrac{d^2}{4}$

$v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})$

Put the value into the formula

$v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}$

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The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}$

Where, $h_{1}=h_{2}$

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$P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)$

Put the value into the formula

$P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)$

$P_{1}=188081.702\ Pa$

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Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

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3 years ago

Which of the following would be used in luminosity calculations?

Daniel [21]

Answer:

1) joule

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3) $10\%$

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This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

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Where:

$V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2}$ is the experimental value

$V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2}$ is the accepted value

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4 years ago

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The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

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Therefore the work done is 17673 J.

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$P=\frac{W}{t}$

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$\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}$

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1 year ago

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kvv77 [185]

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