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1 year ago

7

Look at the variables for this lab. Which variable would we have to change in order to change the amount of current flowing thro

ugh the wire?

1 answer:

____ [38]1 year ago

3 0

Where are the variables from the lab?

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Which uses motion of water in nature to produce electricity?

Zinaida [17]

Answer: D. Hydroelectric power

Explanation: Hydroelectric means water power (not exactly but "hydro" is water and electric is well self explanatory)

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2 years ago

What nonliving materials are important to living things?

Pavel [41]

<span>Nonliving things also have unlimited duration of existence. While living things die and decompose, nonliving things such as rocks, mountains, air and water have existed for millions of years. They may grow, but they do so only by accretion, which is the process of growth by accumulating added layers of matter.</span>

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2 years ago

Read 2 more answers

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

2 years ago

If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years

OLEGan [10]

20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.

5 0

2 years ago

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

vesna_86 [32]

Answer:

$6.63\times 10^8\ N/C$

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

7 0

2 years ago