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Genrish500 [490]
1 year ago
11

Two positive charge spheres, the spheres are separated by 0.40 m. The charge on the first sphere is 100 microcoulombs and the

Physics
1 answer:
agasfer [191]1 year ago
8 0

Answer: 168.75 N

Explanation:

first, let's convert microcoulombs to coulombs

q1 = 1e-4 C

q2 = 3e-5 C

r = 0.4 m

then use the equation Fe = \frac{kq_{1} q_{2}}{r^{2} }

plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2

F = 168.75 N

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Calculate the energy transferred by an appliance using mains electricity (230V) if the charge is 150C. Give your answer in kiloj
Otrada [13]

The energy transferred by the appliance using mains electricity is 17.3 KJ

<h3>Data obtained from the question </h3>
  • Potential difference (V) = 230V
  • Charge (Q) = 150 C
  • Energy (E) =?

<h3>How to determine the energy transferred </h3>

The energy transferred can be obtained as follow:

E = ½QV

E = ½ × 150 × 230

E = 75 × 230

E = 17250 J

Divide by 1000 to express in kilojoules

E = 17250 / 1000

E = 17.3 KJ

Learn more about energy stored in a capacitor:

brainly.com/question/14739936

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3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

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3 years ago
Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT
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PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\\ \tt\bull\leadsto PV

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\\ \tt\bull\leadsto [ML^5T^{-2}]

RHS

\\ \tt\bull\leadsto nRT

\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]

\\ \tt\bull\leadsto [ML^5T^{-2}]

LHS=RHS

hence verified

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It's either A or B because it starts off as nuclear energy.
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