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3 years ago

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A charge Q accumulates on the hollow metallic dome, of radius R, of a Van de Graaff generator. A probe measures the electric fie

ld strength at various points outside the sphere surface. By what factor will the electric field value at the 2R distance be changed if the charge value were increased to (4/3)Q?

2 answers:

timama [110]3 years ago

8 0

Answer:

The factor must be 4/3

Explanation:

the solution is in the attached Word file

frez [133]3 years ago

7 0

Answer:

E'=(3/16)E

Explanation:

The electric field generated by a Van de Graff generator can be taken as the electric field generated by a hollow shell (for distance out of the shell):

$E=k\frac{Q}{R^2}$

Q: electric charge

K: Coulomb's constant

R: distance in which E is measured.

If the distance is increased to R'=2R, and the charge to Q'=(4/3)Q the new electric field is:

$E'=k\frac{Q'}{R'^2}=k\frac{(3/4)Q}{(2R)^2}=\frac{3}{4}k\frac{Q}{4R^2}=\frac{3}{16}k\frac{Q}{R^2}=\frac{3}{16}E$

hence, the new electric field E' is 3/16 times the previous electric field E

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(a) As you ride on a Ferris wheel, your apparent weight is different at the top and at the bottom. Explain. (b) Calculate your a

otez555 [7]

Answer:

a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)

b. Apparent weight at the top is approximately 519.06 N

Apparent weight at the bottom is approximately 558.94 N

Explanation:

a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly

b. The given parameters are

The radius of the Ferris wheel, r = 7.2 m

The period for one complete revolution, t = 28 seconds

The angle covered in one revolution, θ = 2·π radian

The mass of the person riding on the Ferris wheel, the passenger = 55 kg

Therefore, we have;

The angular speed, ω = Δθ/Δt = 2·π/(28)

From which we have;

Centripetal force, $F_c$ = m × ω² × r

Substituting the known values, we have $F_c$ = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N

The centripetal force, $F_c$ = 19.94 N always acting outward from the center

Weight = Mass × Acceleration due to gravity

The weight of the passenger = 55 kg × 9.8 m/s² = 539 N

The weight of the passenger = 539 N always acting downwards

At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards

Therefore;

The net force, which is the apparent weight of the passenger at the top $F_{NET_{Top}}$ = 539 N - 19.94 N ≈ 519.06 N

Apparent weight at the top ≈ 519.06 N

At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards

Therefore;

The net force at the bottom, which is the apparent weight of the passenger at the bottom $F_{NET_{bottom}}$ = 539 N + 19.94 N ≈ 558.94 N

Apparent weight at the bottom ≈ 558.94 N.

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3 years ago

What do the movements of stars and galaxies tell astronomers about how the universe formed?

natulia [17]

This could be Hubble's law, or something related to it. I think there's a possibly Doppler RED SHIFT in the optical spectra of stars etc as observed on the earth. It seems that they are accelerating away from the earth, and that the further away they are the faster they are moving.
It seems that this has been connected to the idea of "The Big Bang" theory of the origin of the universe which seems to have superceded Professor Sir Fred Hoye's Steady State theory of the universe.
There's some Special Relativity in this lot, too.

3 0

3 years ago

) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe

pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0

3 years ago

Read 2 more answers

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

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3 years ago

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bearhunter [10]

Answer:

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3 years ago