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natulia [17]
3 years ago
5

PLEASE PLEASE HELP

Physics
1 answer:
pav-90 [236]3 years ago
6 0

if in series one lightbulb burns out the rest are unable to turn on.

In parallel a single light bulb burns out any other light bulbs are able to work.

Parallel is the best to use during holidays.

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Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out ov
Mumz [18]

Answer:

The number of excess electrons are on the negative surface is 4.80\times10^{10}\ electrons

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}

C=0.285\times10^{-10}\ F

We need to calculate the potential

Using formula of potential

V=Ed

Put the value into the formula

V=18000\times1.5\times10^{-2}\ V

V=270\ V

We need to calculate the charge

Using formula of charge

Q=CV

Put the value into the formula

Q=0.285\times10^{-10}\times270

Q=76.95\times10^{-10}\ C

Here, the charge on both the positive and negative  plates

Q=+76.95\times10^{-10}\ C

Q=-76.95\times10^{-10}\ C

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

n=\dfrac{q}{e}

Put the value into the formula

n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}

n=4.80\times10^{10}\ electrons

Hence, The number of excess electrons are on the negative surface is 4.80\times10^{10}\ electrons

8 0
3 years ago
Write the equation for speed
zzz [600]

Answer:

Distance / Time = Speed

Explanation:

This is because you want to see how much distance you cover within a time interval.

For example, MPH, is a unit for speed, which is miles (distance) per hour (time)

7 0
4 years ago
A capacitor is charged to a potential difference of 12v it delivers 40% of its stored energy to a lamp what is the final potenti
Korolek [52]
Call the capacitance C. 
<span>Note the energy in a capacitor with voltage V is E =½CV². </span>

<span>Initial energy = ½C(12)² = 72C </span>

<span>40% of energy is delivered, so 60% remains.in the capacitor. </span>
<span>Remaining energy = (60/100) x 72C =43.2C </span>

<span>If the final potential difference is X, the energy stored is ½CX² </span>
<span>½CX² = 43.2C </span>
<span>X² = 2 x 43.2 = 86.4 </span>
<span>X = 9.3V</span>
6 0
3 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
elena-s [515]

Answer:

g=13.42\frac{m}{s^2}

Explanation:

1) Notation and info given

\rho_{center}=13000 \frac{kg}{m^3} represent the density at the center of the planet

\rho_{surface}=2100 \frac{kg}{m^3} represent the densisty at the surface of the planet

r represent the radius

r_{earth}=6.371x10^{6}m represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of  the radius

r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}

r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}

So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

So then our linear model would be

\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be dV=r^2 sin\theta d\phi d\theta dr.

And the total mass would be given by the following integral

M=\int \rho (r) dV

Replacing dV we have the following result:

M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

And replacing the values we got:

M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

And now that for any shape the gravitational acceleration is given by:

g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}

4 0
3 years ago
Current X is 2.5 A and runs for 39 seconds. Current Y is 3.8 A and runs for 24 seconds. Which current delivered more charge, and
Aleonysh [2.5K]

Answer: B. Current x delivered 6.3 C more then Y

Explanation:

7 0
4 years ago
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