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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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An object moving 20 m/s
yan [13]

Answer:

<u>We are given:</u>

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

<u />

<u>Solving for Displacement:</u>

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

7 0
3 years ago
A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0
2 years ago
A snowball with a mass of 85 g hits the top hat of a 1.5 m tall snowman and sticks to it. the hat and the snowball, with a combi
LenaWriter [7]
Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,

Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.

v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s

Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2

Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds

Then,
x = 8*0.553 = 4.424 m
7 0
3 years ago
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0
3 years ago
An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligibl
krok68 [10]

(a) The moment of inertia of the wheel  is 78.2 kgm².

(b) The mass (in kg) of the wheel is 1,436.2 kg.

(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

<h3>Moment of inertia of the wheel</h3>

Apply principle of conservation of angular momentum;

Fr = Iα

where;

  • F is applied force
  • r is radius of the cylinder
  • α is angular acceleration
  • I is moment of inertia

I = Fr/α

I = (200 x 0.33) / (0.844)

I = 78.2 kgm²

<h3>Mass of the wheel</h3>

I = ¹/₂MR²

where;

  • M is mass of the solid cylinder
  • R is radius of the solid cylinder
  • I is moment of inertia of the solid cylinder

2I = MR²

M = 2I/R²

M = (2 x 78.2) / (0.33²)

M = 1,436.2 kg

<h3>Angular speed of the wheel after 4 seconds</h3>

ω = αt

ω = 0.844 x 4

ω = 3.376 rad/s

Thus, the moment of inertia of the wheel  is 78.2 kgm².

The mass (in kg) of the wheel is 1,436.2 kg.

The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

Learn more about moment of inertia here: brainly.com/question/14839816

#SPJ1

7 0
1 year ago
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