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KiRa [710]
2 years ago
15

The latent heat of fusion of a substance is the amount of energy associated

Physics
1 answer:
kicyunya [14]2 years ago
6 0

Answer:

The Answer is A Hope I helped you :D Have a Great Day!

Explanation:

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An atom of element X has one more shell of electrons than an atom of beryllium, but it has one less valence electron than beryll
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The correct answer is Sodium
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3 years ago
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Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to
katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

OB=0.40 km

And the Max distance towards bookstore is,

OA=3.65 km

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

AB=\sqrt{OB^{2}+OA^{2}}

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km

Therefore the distance between there destination is 3.67 km.

6 0
3 years ago
If a wave had a wave speed of 1000 m/s and a frequency of 500 Hz, what is its wavelength?
Reil [10]

Answer:

<h2>2 meters</h2>

Explanation:

<h2>Wavelength = Speed/Frequency </h2><h2>1000 m/s ÷ 500 hz </h2><h2>2 m</h2><h2>hz = s</h2><h2>Hopes this helps. Mark as brainlest plz!</h2>

7 0
3 years ago
Sound waves are classified as which type of wave?answer
iren [92.7K]
The answer is D. I hope this helps
4 0
2 years ago
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

  • Q, the charge stored in the capacitor, and
  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

  • \epsilon is the permittivity of the material between the two plates.
  • A is the area of each of the two plates.
  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

V = \displaystyle \frac{Q}{C}.

The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

3 0
3 years ago
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