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IgorLugansk [536]
1 year ago
10

The charge per unit length on a long, straight filament is -90.0μC/m . Find the electric field (b) 20.0cm

Physics
1 answer:
Anna71 [15]1 year ago
5 0

The electric field for 20 cm is 8.09 MN/C

Length of the straight filament = (20cm) (1m/100cm) = 0.2

Electric field for the straight conductor = E = \frac{2k_{e} \lambda}{r}

E = [2(8.99 × 10^9 N·m^2/C^2)(90.0 × 10^-6 C/m)] / 0.2 m

E = 8.09 MN/C

The electric field is directly radially inward, toward the filament.

Electric field is a force produced by a charge near its surroundings. This force is exerted on other charges when brought in the vicinity of this field. SI unit of electric field is N/C (Force/Charge).

Learn more about electric field here:

brainly.com/question/8971780

#SPJ4

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Answer:

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3 years ago
A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar ec
Sonja [21]
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Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.

2) </span><span>anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.

3) </span><span>during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
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4) </span><span>a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.

5) </span><span>a partial lunar eclipse begins when the moon first touches earth's --> 
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6) </span><span> a point at which the moon crosses earth's orbital plane is called a(n)  --> 
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5 0
3 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

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A light bulb converts electrical energy into light energy and also some thermal energy. The amount of electrical energy is 50 Jo
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If the amount of electrical energy is 50 Joules before the conversions, then it would be 50 Joules after the conversion.

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Hope this helps!
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