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just olya [345]
3 years ago
7

A single point charge q is located at the center of both an imaginary cube and an imaginary sphere. How does the electric flux t

hrough the surface of the cube compare to that through the surface of the sphere?
Physics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

Explanation:

Electric flux is defined as the flow of electric field intensity through a given surface.

Mathematically:

\phi=E.A.cos\ \theta

where:

E = electric field

A = area

θ = angle between the area vector and the electric field

Electric flux through the surface of a sphere will be uniform throughout the surface area due to a charge at the center of the sphere. The distance of the surface from the center is always at a constant distance of radius of the sphere.

Electric flux through the surface of a cube will be varying as the surface area is at a varying distance from the center of the cube. The distance of the surface from the center is not at a uniform distance from the center of the cube and so the projection of solid angle changes.

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Can you answer this math homework? Please!
kap26 [50]

\large \mathfrak{Solution : }

9. An object which is in circular motion (moving along a circle) is said to be accelerating because it changes it's direction constantly even if it is moving with a constant speed. cuz acceleration is change in either magnitude or direction of an object with respect to time.

therefore, it's still acceleration as change in direction with time.

10. Average speed of an object can be calculated by dividing the total distance covered by an object by time taken to cover that distance.

i.e

  • \boxed{speed =  \dfrac{distance}{time} }

it can be re- arranged to find the distance as :

  • \boxed{distance = speed \times time}

  • time =  \dfrac{distance}{speed}

11. speed = 20 m/s : conversion into km/h

distance covered : 4 km = 4000 m

  • time =  \dfrac{distance}{speed}

  • t =  \dfrac{4000}{20}

  • t =  200 \:  \: sec

time taken = 200 seconds

12. let's use the first equation of motion to find the acceleration :

  • v = u + at

  • 50 = 80 + 120a

  • 50 - 80 = 120a

  • a =  \dfrac{ - 30}{120}

  • a = -   \dfrac{1}{4} m/s {}^{2}

  • - 0.25 \:  \: m/s {}^{2}
3 0
3 years ago
Diagram of an atom with labels
lina2011 [118]

See this. I hope you find your answer

6 0
3 years ago
In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

8 0
3 years ago
Determine the average acceleration for x(t)=19t^2+7t^3 for a time interval between 3 and 9 seconds
AleksandrR [38]

Answer:

290

Explanation:

Average acceleration is the change in velocity over change in time.

First, find the velocity by taking the derivative of position.

v(t) = dx/dt

v(t) = 38t + 21t²

At t = 3 and t = 9:

v(3) = 303

v(9) = 2043

So the average acceleration is:

a = Δv / Δt

a = (2043 − 303) / (9 − 3)

a = 290

Use appropriate units.

7 0
3 years ago
Name two factors that can affect the function of an enzyme
myrzilka [38]

Answer:

1. pH

2. Temperature

Hope this helps

3 0
3 years ago
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