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2 years ago

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A 1.0 m long piece of coaxial cable has a wire with a radius of 1.1 mm and a concentric conductor with inner radius 1.3 mm. The

area between the cable and the conductor is filled with a dielectric. If the voltage drop across the capacitor is 6000 V when the line charge density is 8.8 μC/m, find the value of the dielectric constant. (k = 1/4πε₀ = 8.99 × 109 N · m²/C²)A) 4.8
B) 5.3
C) 4.4
D) 5.7

1 answer:

Alex Ar [27]2 years ago

5 0

Answer:

C) 4.4

Explanation:

The potential of a cylindrical capacitor is given by the formula:

$V=\frac{2kq}{L\epsilon}ln(\frac{a}{b})\\\\\epsilon=\frac{2kq}{LV}ln(\frac{a}{b})$

where:

k : Coulomb Constant

L : length of the capacitor

a : outer radius

b : inner radius

V : potential

By replacing we obtain:

$\epsilon=\frac{2(8.89*10^{9}N/m^2C^2)(8.8*10^{-6}C)}{(1m)(6000V)}ln(\frac{1.3mm}{1.1mm})=4.35$

Hence, the answer is C) 4.4 (4.35 is approximately 4.4)

hope this helps!!

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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2 years ago

A pump lifts water from a lake to a large tank 20 m above the lake. How much work against gravity does the pump do as it transfe

Aleonysh [2.5K]

Answer:

980 kJ

Explanation:

Work = change in energy

W = mgh

W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)

W = 980,000 J

W = 980 kJ

The pump does 980 kJ of work.

3 0

2 years ago

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Small frogs that are good jumpers are capable of remarkable accelerations. One species reaches a takeoff speed of3.7 m/s in60 ms

MAVERICK [17]

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = $6 *10^{-3} seconds$

So acceleration of frog = $\frac{3.7}{60*10^{-3}}$

= 61.66 m/ $s^2$

So acceleration of frog is 61.66 m/ $s^2$

o it is evident that frog is capable of remarkable accelerations.

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2 years ago

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Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

So

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

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2 years ago

the distance between the sun and earth is about 1.5X10^11 m. express this distance with an SI prefix and kilometers

Angelina_Jolie [31]

First, we write the SI prefixed. The SI unit for distance is meters.

Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²

Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.

1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm

Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.

1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km

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2 years ago