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vladimir1956 [14]
3 years ago
7

A source charge of 5.0 µC generates an electric field of 3.93 × 105 at the location of a test charge. How far is the test charge

from the source charge?
Physics
2 answers:
Nonamiya [84]3 years ago
8 0

Answer:

0.34 m

Explanation:

Just did assignment

kap26 [50]3 years ago
6 0

Answer:

B. )  0.34 m

I definitely guessed and got the right answer so :))

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20 POINTS
Olegator [25]

Answer:

I believe the answer is B.

Explanation:

Newton's First Law of Gravity states, "The greater the weight (or mass) of an object, the more inertia it has. Heavy objects are harder to move than light ones because they have more inertia. "

6 0
3 years ago
Read 2 more answers
When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of
Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0
3 years ago
A dielectric material such as paper is inserted between the plates of a capacitor as the capacitor holds a fixed charge on its p
Vikki [24]

Answer:

Introducing a dielectric into a capacitor decreases the electric field, which decreases the voltage, which increases the capacitance.

Explanation:

A dielectric (or dielectric material) is an electrical insulator that can be polarized by an applied electric field. When a dielectric material is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor but only slightly shift from their average equilibrium positions causing dielectric polarization

Types of dielectric material

Ceramic, Mica paper glass

7 0
3 years ago
The base of a pyramid covers an area of 13.0 acres (1 acre = 43,560 ft2) and has a height of 481 ft. If the volume of a pyramid
Allisa [31]
V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
3 0
3 years ago
Read 2 more answers
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
2 years ago
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