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3 years ago

What does the sun consume to produce energy and helium?

Physics

2 answers:

Art [367]3 years ago

6 0

Answer:

Sun consumes Hydrogen to produce energy

Explanation:

Sun is a huge sphere of hydrogen gas in which hydrogen gas nuclei combines and forms helium nuclei by nuclear fusion reaction.

This is also known given as

$_1^2H + _1^3H --> _2^4He + _0^1n + energy$

so here when these two isotopes of hydrogen combines and gives helium nuclei then in this reaction the energy is released and due to this nuclear energy released the sun will get energy to consume and to give high temperature.

aev [14]3 years ago

3 0

This reaction, known as nuclear fusion, converts hydrogen atoms into helium. The by-product of nuclear fusion in theSun's core is a massive volume ofenergy that gets released and radiates outward toward the surface of the Sunand then into the solar system beyond it.

just reword it

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At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

4 years ago

The graph below represents the relationship between the force exerted on an elevator and the distance the elevator is lifted.

Stella [2.4K]

I have three problems with this question.

#1). If you copied the question exactly the way it's written,
then the question is written very badly. The wording is
misleading, and the more you try to think about it and
puzzle it out, the more it'll damage your understanding
of Physics.

There is no relationship between the force exerted on an
elevator and the distance the elevator is lifted.

-- If the force is anything more than the weight of the elevator ...
even one ounce more ... then it'll lift the elevator as high as
you want.

-- If the force is anything less than the weight of the elevator ...
even one ounce less, then that elevator is headed for the bottom.

#2). You didn't post any graph below, so if we need the graph
to answer the question, then we can't answer the question.

#3). I guess that's OK, because you didn't ask any question.

8 0

3 years ago

A 5000kg car traveling at 40m/s crashes into a wall and comes to a complete stop in 10ms. What was the force on the wall?

Kisachek [45]

Answer:

20,000,000 N

Explanation:

First find the acceleration:

a = Δv / Δt

a = (0 − 40 m/s) / 0.010 s

a = -4000 m/s²

Next use Newton's second law to find the force on the car:

F = ma

F = (5000 kg) (-4000 m/s²)

F = -20,000,000 N

According to Newton's third law, the force on the wall is equal and opposite the force on the car.

F = 20,000,000 N

7 0

3 years ago

Calculate the centripetal force acting on a 925 kg car as it rounds an unbanked curve with a radius of 75 m at a speed of 22 m/s

igomit [66]

Since f=m(v^2/r),or fnet is equal to ma.

force = unknown
velocity=22m/s
radius=75m

f=m(v^2/r)
f=925(22^2/75)
f=5969.333N

3 0

3 years ago

Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water le

mafiozo [28]

Answer:

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

Explanation:

Given;

Diameter of hose d = 2.76 cm

Volume filled V = 20.0 L = 20,000 cm^3

Time t = 1.45 min = 105 seconds

The volumetric flow rate of water is;

F = V/t = 20,000cm^3 ÷ 105 seconds

F = 190.48 cm^3/s

The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.

F = Av

v = F/A

Area A = πd^2/4

Speed v = F/(πd^2/4)

v = 4F/πd^2 ......1

Substituting the given values;

v = (4×190.48)/(π×2.76^2)

v = 31.83767439628 cm/s

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

3 0

3 years ago