PHYSICS COLLEGE

Assume the electric field has a magnitude of 59.0 m V , and the particle has a net positive charge of Q=+5.40 C. Calculate the magnitude of the electric potential difference, \vert \Delta V \vert = \vert V_B - V_A \vert∣ΔV∣=∣V B −V A ∣.

Answers

Answer 1

Answer:

V = 1.69 * 10^6 V

Explanation:

Parameters given:

Electric field, E = 59V/m

Charge, q = 5.40C

We need to first find the distance between the electric charge and the point of consideration to be able to find the Electric potential difference.

Electric field is given as:

E = (kq/r^2)

k = Coulombs constant

=> r^2 = kq/E

=> r^2 = (9 * 10^9 * 5.4) / 59

r^2 = 8.2 * 10^8

r = 2.84 * 10^4 m

We can now find the Electric Potential by using:

V = kq/r

Hence,

V = (9 * 10^9 * 5.4) / (2.84 * 10^4)

V = 1.69 * 10^6 V

Related Questions

Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.
1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?
A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?
Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

Answers

Answer:

The expression would be ω = $\sqrt{(g)/(L sin 0 ) }$

Explanation:

Given that ω is the angular velocity

g is the acceleration due to gravity

L is the length

θ is the angle of downward tilt

For an object we compare the horizontal and vertical component of the forces acting on the body;

For vertical component

T sinθ = mg............1

For the horizontal component

T cos θ = $(mv^(2) )/(R)$ .............2

R is our radius and is = L cos θ

v = ωR

substituting into equation 2 we have

T cos θ = m(ωR $)^(2)$ /R

T cos θ=m(ω $)^(2)$ R ..................3

Now comparing the vertical and the horizontal component we have;

equation 1 divided by equation 3 we have

T sin θ /T cos θ = mg / m(ω $)^(2)$ R

Tan θ = g / (ω $)^(2)$ R............4

Making ω the subject formula we have;

(ω $)^(2)$ = g/ R Tan θ

But R = L cos θ and Tan θ = sin θ/ cosθ

putting into equation 4 we have;

(ω $)^(2)$ = g /[( L cos θ) x( sin θ/ cosθ)]

(ω $)^(2)$ = g/ L sinθ

ω = $\sqrt{(g)/(L sin 0 ) }$

Therefor the expression for the angular velocity ω in terms of g, L and angle θ would be ω = $\sqrt{(g)/(L sin 0 ) }$

Let g, r, L and T are gravity, radius, length, and angle of string w/r/t vertical, respectively.
Then
ω²r = rg/Lcos T
ω² = g/L cos T
ω = √(g / L cos T)

Explain what happent to the pressure exerted by an object when the area over which it is exerted:a) increase
b) decrease

Answers

The pressure decreases if the area increases. If the area decreases then the pressure increases.

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off, it continues to rotate with a constant angular deceleration for 10.0s before coming to rest.Part A

If its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )

Part B

How many revolutions did the centrifuge complete after being turned off?

Answers

Answer:

$a_r=389\ rev.s^(-2)$

$n=58350 rev$

Explanation:

Given:

time of constant deceleration, $t=10\ s$

initial angular speed, $N_i=3890\ rpm\$

Using equation of motion:

$N_f=N_i+a_r.t$

$0=3890+a_r* 10$

$a_r=389\ rev.s^(-2)$

Using eq. of motion for no. of revolutions, we have:

$n=N_i.t+(1)/(2) a_r.t^2$

$n=3890* 10+0.5* 389* 100$

$n=58350\ rev$

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.

Answers

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

Electrons with energy of 25 eV have a wavelength of ~0.25 nm. If we send these electrons through the same two slits (d = 0.16 mm) we use to produce a visible light interference pattern what is the spacing (in micrometer) between maxima on a screen 3.3 m away?

Answers

Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

$\beta=(\lambda D)/(d)$

Put the value into the formula

$\beta=(0.25*10^(-9)*3.3)/(0.16*10^(-3))$

$\beta=5.15*10^(-6)\ m$

$\beta=5.15\ \mu m$

Hence, The spacing is 5.15 μm.

Final answer:

To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.

Explanation:

To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.

Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.

x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm

So, the spacing between maxima on the screen is approximately 5.14 micrometers.