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2 years ago

What is the phase of a 20 gram of crystalline when its temperature is 250 degrees clesius

Physics

1 answer:

loris [4]2 years ago

5 0

Explanation:

20÷250 =12.5 and do it by using fourmulaes

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Ariel's puppy weighs 15 lb. Heather's puppy weighs 50 N. Whose puppy weighs more?

yarga [219]

15 lb is nearly 6 kg amd 50 n is nearly 5 kg,so first puppy weighs more

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3 years ago

Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the conta

hodyreva [135]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file has a detailed solution of the given problem.

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3 years ago

Your friend turns the volume up on their speaker system, so that the sound intensity is 4 times greater than it was beforehand.

erica [24]

Answer:

Explanation:

We know that intensity I = P/A where P = power and A = area through which the power passes through.

Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.

So, I = P/A and I₀ = P₀/A

Now, if I = 4I₀,

P/A = 4P₀/A

P = 4P₀

Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t

Substituting P and P₀ into the equation, we have

P = 4P₀

E/t = 4E₀/t

E = 4E₀

Since the energy is four times the initial energy, the energy output increases by a factor of 4.

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3 years ago

If the action force is a player kicking a soccer ball, then what is the reaction force?

LiRa [457]

The force acting on his feet.

4 0

3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$