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3 years ago

Please answer. give full explanation.

Physics

1 answer:

Ad libitum [116K]3 years ago

4 0

Work done=F *s*cos theta
Work done in this case can only be in the direction in which the electrons move.Thus cos theta is multiply to find the component of the force ( acting on an object doing work) which acts in the direction of propagation on movement of the object.Anyway,as in this case ,as magnetic field act only perpendicularly (at 90°) to electric fields(those are the electrons)no work is done.This is cause at 90°,there is no component of the magnetic field exiting of force in the direction of propagation of the electric field (electrons),as cos of 90 ° is 0, no work is done by the magnetic field.(as the whole product becomes zero regardless of the force or the displacement/distance

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For a 1 efficient step down transformer voltage in primary and secondary are equal current in primary and secondary are equal in

Fynjy0 [20]

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<h3>What Does a Step-Down Transformer Mean by Power?</h3>

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2 years ago

What is the difference between velocity and acceleration and how do we find each one

katen-ka-za [31]

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6 0

3 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

4 years ago

A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed

Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0

3 years ago

A basketball is thrown horizontally with an initial speed of

storchak [24]

Answer:

1.35 m

Explanation:

Taking down to be positive, given:

Δx = Δy / tan 30.0º

v₀ₓ = 4.50 m/s

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = 10 m/s²

Find: Δy

First, find the time it takes to land in terms of Δy.

Δy = v₀ t + ½ at²

Δy = (0 m/s) t + ½ (10 m/s) t²

Δy = 5t²

Next, find Δx in terms of t.

Δx = v₀ t + ½ at²

Δx = (4.50 m/s) t + ½ (0 m/s) t²

Δx = 4.50t

Substitute:

Δy = 5 (Δx / 4.50)²

20.25 Δy = 5 (Δx)²

4.05 Δy = (Δx)²

4.05 Δy = (Δy / tan 30.0º)²

4.05 Δy = 3 (Δy)²

1.35 = Δy

The basketball was thrown from an initial height of 1.35 m.

Graph: desmos.com/calculator/ujuzdo9xpr

8 0

3 years ago