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Umnica [9.8K]
3 years ago
15

Which of the following is an example of a primary source?

Physics
1 answer:
Oduvanchick [21]3 years ago
8 0

Answer:

i think its no c.

a textbook written by a scientist. because its a first hand data.

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1. write the meaning of the following terms:electrostatic,neutral, positively charged, negatively charged, coulomb,microcoulomb,
dybincka [34]

ELECTROSTATIC:

relating to stationary electric charges or fields as opposed to electric currents.

NEUTRAL:

nor negative nor positive/having no charge

POSITIVELY CHARGED:

positive charge occurs when the number of protons exceeds the number of electrons

NEGATIVELY CHARGED:

negative charge occurs when the number of electrons exceeds the number of protons.

COULOMB:

SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.

MICROCOULOMB:

a unit of electrical charge equal to one millionth of a coulomb.

NANOCOULOMB:

Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.

CONSERVATION OF CHARGE:

constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction

QUANTISATION OF CHARGE:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

5 0
2 years ago
A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block attached to it by 0.5 m. The block is released from
morpeh [17]

Answer:

0.7000cm

Explanation:

bbbgffffffjj office yhhjujhhhhhhhhhhhjjnjhhhhbhhh

6 0
3 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8
pashok25 [27]
Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

6 0
3 years ago
You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart
hichkok12 [17]

Answer:

v = 3.5 \times 10^5 m/s

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}

now we have

\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}

\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}

so we will have

r = 3.46 \times 10^8 m

Now by energy conservation

-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}

-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5

v = 3.5 \times 10^5 m/s

7 0
3 years ago
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