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2 years ago

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Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p

late. What is the ratio U1/U2 of their potential energies?

1 answer:

Svetlanka [38]2 years ago

5 0

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

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A sports car accelerates from 0 to 30 mph in 1.5 s. How long would it take to accelerate from 0 to 60 mph, assuming the power of

Crank

Answer:

6 s

Explanation:

given,

Sports car accelerate from 0 to 30 mph in 1.5 s

time taken to accelerate 0 to 60 mph = ?

The power of the engine is independent of velocity and neglecting friction

power =

P = constant

the kinetic energy for 60 mph larger than this of 30 mph

= $\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}$

= $\dfrac{v_1^2}{v_2^2}$

= $\dfrac{60^2}{30^2}$

= 4

gain in kinetic energy = P x t

time = 4 x 1.5

= 6 s

8 0

1 year ago

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount

Mariulka [41]

Answer:

The value is $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

The radius of the sphere is $r = 6.32 *10^{8} \ m$

The temperature is $T_x = 5350 \ K$

The average temperature of the rest of the universe is $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

$\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

$s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

$Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value

$\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=> $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=> $Q = 2.332 *10^{26} \ J$

So

$s_r = \frac{2.332 *10^{26}}{2.73}$

=> $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

$s_x = \frac{Q}{T_x}$

=> $s_x = \frac{2.332 *10^{26} }{5350}$

=> $s_x = 4.359 *10^{22} \ J/K$

So

$\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=> $\Delta s = 8.537 *10^{25 } \ J/K$

7 0

1 year ago

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25

antiseptic1488 [7]

we are given in the problem the following dimensions or specifications
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7

The formula that is applicable from physics is
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes

7 0

1 year ago

Read 2 more answers

If the average speed of an orbiting space shuttle is 27 800 km/h, determine the time required for it to circle Earth. Assume tha

balu736 [363]

Answer:

t = 1.51 hours

Explanation:

given,

Speed of space shuttle. v = 27800 Km/h

Radius of earth, R = 6380 Km

height of shuttle above earth, h = 320 Km

Total radius of the shuttle orbit

r' = R + h

r' = 6380 + 320

r' = 6700 Km

distance, d = 2 π r

d = 2 π x 6700

$time = \dfrac{distance}{speed}$

$time = \dfrac{2\pi\times 6700}{27800}$

t = 1.51 hours

Time require by the shuttle to circle the earth is equal to 1.51 hr.

7 0

1 year ago

The 600-N ball shown is suspended on a string AB and rests against the frictionless vertical wall. The string makes an angle of

Harrizon [31]

Answer: T = 692.82 and 346.4 N

Explanation:

Given that;

w = 600 N

∅ = 30°

ΣFy = ma

a = 0 m/s²

ΣF = T(cos30°) - W = 0

T(cos30°) = W

we Divide both sides by cos30°

T = W / cos30o

T= 600N / cos30°

T = 692.82

and ∑fx

F = T sin∅

F = 692.82 × (sin30°)

F = 346.4 N

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1 year ago