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Whitepunk [10]
3 years ago
7

turning on the television is a form of ___ energy that transforms into ___ energy and/or sound energy

Physics
1 answer:
MariettaO [177]3 years ago
8 0
Electric energy into light energy and sound energy
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Which of the following is the most appropriate description of an electric current?
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Electric current is the flow of charge due to the potential difference between two terminals per unit time. It is denoted by I and its unit is amp. It can be mathematically expressed as I=Q/t.

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Derek is watching steam form swirling patterns above the boiling water in his beaker.
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Convection

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Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from re
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Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).

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3 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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