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3 years ago

The law of reflection states that the angle of reflection is equal to the angle of

Physics

1 answer:

Natali [406]3 years ago

3 0

The law of reflection states that the angle of reflection is equal to the angle of Incidence .

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Two point charges lie on the x axis. A charge of + 2.30 pC is at the origin, and a charge of − 4.50 pC is at x=−11.0cm.

Radda [10]

r₁ = distance of point A from charge q₁ = 0.13 m

r₂ = distance of point A from charge q₂ = 0.24 m

r₃ = distance of point A from charge q₃ = 0.13 m

Electric field by charge q₁ at A is given as

E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right

Electric field by charge q₂ at A is given as

E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left

Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction

Electric field at A will be zero when

E₁ = E₂ + E₃

1.225 = 0.703 + E₃

E₃ = 0.522 N/C

Electric field by charge "q₃" is given as

E₃ = k q₃ /r₃²

0.522 = (9 x 10⁹) q₃/(0.13)²

q₃ = 0.980 x 10⁻¹² C = 0.980 pC

4 0

3 years ago

Why might it be important to know your inherited traits passed down by your parents?

ExtremeBDS [4]

Answer:

health factors that could harm you or your offspring

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2 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

Use the galvanometers to determine the amount and direction of the induced current. Which galvanometer is

docker41 [41]

Answer:

Option B

Explanation:

Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.

For galvanometer B, the reading is 4 mA.

While for galvanometer C, the reading is -5 MA

Thus, option B is correct.

4 0

3 years ago

What conditions must be satisfied for a body to be in equilibrium <br>

dem82 [27]

For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero.

5 0

3 years ago