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Ivan
3 years ago
7

Which would show an example of how physical changes are reversible?

Physics
2 answers:
Rina8888 [55]3 years ago
7 0

Answer:

melting tin and then cooling it into a mold

Explanation:

laiz [17]3 years ago
3 0
Examples<span> of </span>reversible changes<span> include melting chocolate and </span>changing<span> it back into a solid by cooling it, and melting candle wax by heating it and solidifying the wax by cooling it. </span>Reversible changes<span> are </span>changes<span> that can be reversed. They are also known as </span>physical changes<span>.</span>
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A block with mass ma = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second
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<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation. F = Ma F is the net force M is the total mass of the system a is the acceleration a = F / M a = (mb)(g) / (ma + mb) a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg) a = 58.8 N / 20 kg a = 2.94 m/s^2 The magnitude of the acceleration of the system is 2.94 m/s^2</span>
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3 years ago
Who speaks the line "Lord, what fools these mortals be"?
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The answer is D.Puck.

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3 0
3 years ago
an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco
Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

a_c = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

6 0
3 years ago
The two main states of mechanical energy are ___________ and potential energy.
shutvik [7]
<span>Potential energy and Kinetic energy</span>
6 0
3 years ago
Read 2 more answers
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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