PHYSICS COLLEGE

An electron is moving in the presence of an electric field of 400 N/C. What force does the electron experience?

Answers

Answer 1

Answer:

Answer:

Explanation:

Given

Electric Field strength $E=400\ N/C$

charge of electron $q=1.6* 10^(-19)\ C$

Force on the charge particle is given by

$F=qE$

$F=1.6* 10^(-19)* 400$

$F=640* 10^(-19)\ N$

but this force will be acting in the direction opposite to the direction of Electric field because electron is negatively charged

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On a map each inch represents 25 miles what is the length of a highway if it is 6 inches long on a map

Answers

Answer:

150 hope this helps

Explanation:

Answer:

150

Explanation:

Answer all three parts and show work.

Answers

The distance for both Parts A and B are given in the question.

A balloon drifts 140m toward the west in 45s.

The wind suddenly changes and the balloon flies 90m toward the east in the next 25s.

To find the total distance, we can just add.

140 + 90 = 230m

Best of Luck!

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

$P_r=(IA)/(c)$

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

$P=140W=140(J)/(s)\n0.4*140J=56J$

56J is the electromagnetic energy.

The area of the bulb is

$A_b=4\pi r^2=4\pi (14m)^2=2463m^2$

The radiation pressure is

$P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa$

hope this helps!!

The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v = $(\Delta x)/(\Delta t)$

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

Learn more about Speed of a Particle here:

brainly.com/question/32295338

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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

The 1 here represented the central bright spot

$n_(max) = 2 * 2500 + 1$

$n_(max) =5001$

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