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3 years ago

Which object has the most thermal energy?

Physics

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

D. A 10 kg rock at 15°C

Explanation:

Thermal energy is heat energy, meaning that the hotter rocks will have more energy.

The heavier/larger rocks can store more energy.

So if we combine the largest two variables, we get a 10kg rock at 15°C.

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When you are high up in the air you<br> have<br> greater potential energy<br> less potential energy

mina [271]

Answer:

Its Greater potential energy because the air is high up and that makes high energy power

Explanation:

5 0

3 years ago

Which provides the best analogy for an electron in an atomic orbital?

Svetllana [295]

<h2>Answer: a bee trying to escape from a closed jar </h2>

In an atom the electrons will occupy orbitals so that their energy is as small as possible. That is why the orbitals are ordered based on their energy level in an increasing order, which is associated with a particular range of energy based on its distance from the atom nucleus.

In this sense, an electron "jumps" from one level to another in the atom in the same way a bee tries to escape from a closed jar.

3 0

2 years ago

________ is a measure of how many waves pass by in one second.

Sergio [31]

C. Frequency is a measure of how many waves pass by in one second.

4 0

3 years ago

A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus

babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$ , $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$ , then the impulse applied by the stick to the park is approximately:

$I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]$

$I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]$

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

8 0

3 years ago

Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so

mote1985 [20]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

4 0

2 years ago