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Veseljchak [2.6K]

3 years ago

10

With energy, changes in the physical world are possible. These can be a change in speed, a change in direction, a change from co

ld to hot, and many other kinds of changes.
Question 4 options:
True
False

2 answers:

Yakvenalex [24]3 years ago

4 0

Answer:

i believe it is true

Explanation:

hope this helps

laiz [17]3 years ago

3 0

True because energy can change its speed because of temperature and also its direction

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What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the <u>electric constant</u>

$q=1.6(10)^{-19}C$ is the <u>electric charge of the hydrogen nucleus</u>, which is positive

$r=5.292(10)^{-11}m$ is the <u>distance</u>

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

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4 years ago

Why are the element from period 2 grouped together

valentinak56 [21]

All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. As you move down the table, every row adds an orbital.

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4 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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4 years ago

A ball is thrown horizontally at 6.0 m/s from a cliff 80 meters high. How far from the base of the cliff will the ball hit the g

UNO [17]

Answer:

160 meters I think

Explanation:

if I'm wrong I'm a troop juSt trying to get points to ask a couple of questions

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Murljashka [212]

Answer:0.000002

Explanation: I Looked It Up lol

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