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3 years ago

Cold water will warm to room temperature faster in a: a) Black pot

Physics

1 answer:

Viktor [21]3 years ago

8 0

Answer:

black pot

Explanation:

option a is correct

Cold water will warm to room temperature faster in a Black pot as we know that Black surfaces and good absorbers of heat in comparison to other surfaces. Moreover, Silver a good reflector of heat. It reflects most of heat incident on it, the cold water will take longer time to come the room temperature in Silver pot.

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What is dark energy?

expeople1 [14]

Dark energy is a theoretical repulsive energy that causes the acceleration to the expansion of the universe. Among the choices, the nearest answer would be D. Dark energy can also be defined as a new form of energy, that is a dynamic field that fills up space but has an effect opposite to that of normal energy.

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3 years ago

If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Cal

Alekssandra [29.7K]

The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

<h3>emf induced in the loop</h3>

The magnitude of e.m.f induced in the loop is calculated as follows;

emf = dФ/dt

Ф = 6t² + 7t

dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

Learn more about emf here: brainly.com/question/24158806

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2 years ago

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

Blababa [14]

The magnitude of charge on a proton and electron is the same, 1.602 x 10-19 C. Protons are +, and electrons -.

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3 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)

PilotLPTM [1.2K]

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

<h3>De Broglie wavelength:</h3>

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

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2 years ago