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lorasvet [3.4K]
3 years ago
10

Which of these statements would best explain the problem encountered with nuclear waste disposal? A) The isotopes have a long ha

lf-life and only remain radioactive for a long time period. B) The isotopes have a short half-life and only remain radioactive for a long time period. C) The isotopes have a long half-life and only remain radioactive for a short time period. D) The isotopes have a short half-life and only remain radioactive for a short time period. 
ANSWER IS A)
Physics
1 answer:
algol133 years ago
4 0
That's right, the correct answer is
<span>A) The isotopes have a long half-life and only remain radioactive for a long time period

The half life of an isotope is the time it takes for the amount of the sample to reduce to half of its initial value. If an isotope has a long half-life, it means it takes a long time to reduce down to a significant level, so it will remain radioactive for a long time period.</span>
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A circuit has a resistance of 8 Ohms. the voltage supplied to the circuit is 14 volts. what is the current flowing through it?
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A woman takes her dog Rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho
ValentinkaMS [17]

Answer:

The force parallel to the horizontal is 26.24 N

Explanation:

She pulls on the leash with a force F = 30 N, this force, since its at an angle of 29° (i will cal this angle \theta), it has a force component on x (the horizontal, i will call this force F_{x}) and a force component on y (the vertical, i will call this F_{y} ).

This can be seen in the attached picture.

Since we are asked about the force parallel to the horizontal, we need to find the component of the force F_{x}, since F_{x} is the adjacent angle, we need to use cosine:

F_{x}=Fcos \theta

since F=30N and \theta=29

F_{x}=(30N)cos(29)

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F_{x}=26.24N

The force parallel to the horizontal is 26.24 N

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3 years ago
"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its
prisoha [69]

Answer:

1.3m/s

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we can express the velocity by the derivative of the displacement,

Hence

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
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Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

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According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

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\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

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Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
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