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3 years ago

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A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell

itself. Calculate the total charge on the outer surface of the conductor.

1 answer:

Lisa [10]3 years ago

7 0

Answer: 7.53 μC

Explanation: In order to explain this problem we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

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A sudden discharge of static electricity during a thunderstorm is called

Tpy6a [65]

Hello
It is called lightning. Lightning in a storm occurs when there are two regions (it can be cloud-cloud or cloud-ground), one with a strong excess of positive charges and the other one with a strong excess of negative charges. The two types of charge attract each other, and then a sudden flow of charges from one region to the other occurs, which is called lightning.

3 0

3 years ago

How much electrical energy is used by the river cooker oven rated 800 W ,230 V is switched on for 30 minutes

mariarad [96]

Answer:

E = 1440 kJ

Explanation:

It is given that,

Power of a cooker oven is 800 W

Voltage at which it is operated is 230 V

Time, t = 30 minutes = 1800 seconds

We need to find the electrical energy used by the cooker oven. The product of power and time is equal to the energy consumed. So,

$E=P\times t\\\\E=800\ W\times 1800\ s\\\\E=1440000\ J\\\\\text{or}\\\\E=1440\ kJ$

So, electrical energy of 1440 kJ is consumed by the cooker oven.

8 0

3 years ago

The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con

ira [324]

The Period of the resulting shm will be T=39.7

<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

ω=k/m

ω=1200/3

ω=400

<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the torque value </u>

τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

<u>Rotational to linear form</u>

a = αr

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

6 0

3 years ago

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.

CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm$

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the heigth of the image

$h_o$ is the height of the object

$q=36.0 cm$

$p=16.0 cm$

Solving the formula for $h_i$ , we find

$h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm$

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of $h_i$ is positive, the image is erect

- When the sign of $h_i$ is negative, the image is inverted

In this case, $h_i$ is negative, so the image is inverted.

4 0

3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

2 years ago