3 years ago

A star produces energy by? ( a p e x )

Physics

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer and Explanation:

Stars create energy primarily through the fusion of hydrogen into helium through nuclear reactions.

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Which organelle acts like the transportation or circulatory system of the cell?

REY [17]

The endoplasmic rectiulum... hope this helps!

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3 years ago

Which is more dense 10g of shampoo or 10kg of the same shampoo

aleksandr82 [10.1K]

Density is a property of the substance, and doesn't depend on
how much of it you have.

One drop of the shampoo and a supertanker-full of the same
shampoo have the same density.

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3 years ago

A circuit has a current of 3.6 A and a resistance of 5.0 Ω. What is the voltage applied to the circuit?

Inessa [10]

The key formula ===> Voltage = (current) x (resistance)

Plug in the numbers given ===> Voltage = (3.6 A) x (5.0 ohms)

Dooda multiplication ===> Voltage = 18 volts

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3 years ago

Read 2 more answers

A coil has an inductance of 5 H and a resistance of 20 Ω. If a DC voltage of 100 V is applied to the coil, find the energy store

nadya68 [22]

Answer:62.5 J

Explanation:

Given

Inductance(L)=5 H

resistance(R) $=20 \Omega$

Voltage(V)=100 V

$Current=\frac{100}{20}=5 A$

Current in L-R circuit is given by

$I=I_0\left [ 1-e^{-\frac{Rt}{L}}\right ]$

and Power=i^2R+Li\frac{di}{dt}[/tex]

For Steady state i.e. at $t=\infty$

$I=I_0$

Energy Stored is $E=\frac{Li^2}{2}$

$E=\frac{5\times 5^2}{2}=62.5 J$

6 0

3 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

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3 years ago