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2 years ago

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An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 30 mm and 80 mm

. The electric potential of the inner conductor with respect to the outer conductor is -466 V (k = 1/4πε 0 = 8.99 × 109 N · m2/C2). The average energy density of the capacitor is closest to\

1 answer:

Licemer1 [7]2 years ago

4 0

Answer:

24

Explanation:

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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

$34 v + 54 (3 m/s) = 0$

$v = -4.76 m/s$

8 0

2 years ago

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

2 years ago

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging riv

zzz [600]

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

$V_2^2=V_o^2-2*g*\Delta h$

$V_2^2=-2*10*(-31)$

$V_2=24.9m/s$ This will be our initial speed for a balance of energy.

By conservation of energy:

$m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0$

Where

$h$ is your height at its maximum elongation

$h_o$ is the height of the bridge

$l_o$ is the length of the unstretched bungee cord

$800h+21*(79-h)^2-63200-24800.4=0$

$21h^2-2518h+43060.6=0$ Solving for h:

$h_1=20.66m$ and $h_2=99.24m$ Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

6 0

2 years ago

Read 2 more answers

A thin layer of oil with index of refraction no = 1.47 is floating above the water. The index of refraction of water is nw = 1.3

garik1379 [7]

Answer:

A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.

Part (a) Express the wavelength of the light in the oil, λ₀, in terms of λ and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of λ o

(c) Express tmin in terms of λ and no.

(d) Solve for the numerical value of tmin in nm.

Explanation:

n₀ = 1.47

refraction of water = 1.3

refraction of air = 1

wavelength λ = 775 nm

(a) wavelength of light in water ⇒ λ₀ = λ / n₀

(b) minimum thickness of the film that will result in destructive interference

t(min) = λ₀ / 2

(c) the express t(min)

t = λ /2n₀

(d) the thickness is

t = 775 / 2(1.47)

= 263.61 nm

4 0

2 years ago