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2 years ago

10

The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi

eld along the disk's axis at the following distances from its center?

1 answer:

iragen [17]2 years ago

3 0

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

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A 230-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1,100 A. If the conductor is copper

Molodets [167]

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

5 0

2 years ago

Read 2 more answers

A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5

Paul [167]

Answer:

Energy lost due to friction is 22 J

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball $KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}\times 4\times 6^2=72J$

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

$KE=\frac{1}{2}\times 4\times 5^2=50J$

Therefore energy lost due to friction = 72 -50 = 22 J

8 0

2 years ago

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

2 years ago

Skater 1 has a mass of 105.0 kg and a velocity of 2.0 m/s to the left. He

mart [117]

The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

<h3>What is velocity?</h3>

Velocity can be defined as the ratio of the displacement and time of a body.

To calculate the final velocity of Skater 1 we use the formula below.

Formula:

mu+MU = mv+MV............ Equation 1

Where:

m = mass of the first skater
M = mass of the second skater
u = initial velocity of the first skater
U = initial velocity of the second skater
v = final velocity of the first skater
V = final velocity of the second skater.

make v the subject of the equation.

v = (mu+MU-MV)/m................ Equation 2

Note: Let left direction represent negative and right direction represent positive.

From the question,

Given:

m = 105 kg
u = -2 m/s
M = 71 kg
U = 5 m/s
V = -3.4 m/s.

Substitute these values into equation 2

v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
v = (-210+355+241.4)/105
v = 386.4/105
v = 3.68 m/s
v ≈ 3.7 m/s

Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

Learn more about velocity here: brainly.com/question/25749514

7 0

1 year ago

A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn

vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

$f_{m} = \mu_{s}N$

where

N = Normal reaction = mg

Thus

$f_{m} = \mu_{s}mg$

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

$f\leq f_{m}$

$f \leq \mu_{s}mg$

$109 \leq \mu_{s}\times 31.2\times 9.8$

$\mu_{s}\geq 0.356$

7 0

2 years ago