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Ad libitum [116K]
3 years ago
5

Imagine that the light you observe from space is exhibiting a redshift. This would mean that the Universe is

Physics
1 answer:
Ainat [17]3 years ago
7 0
Expanding and galaxies (and stars) are moving further away.
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How does radioactive decay work
Lesechka [4]

Answer:

Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.

Explanation:

6 0
4 years ago
How are slow twitch fibers and fast twitch fibers different?
Vlad1618 [11]
Slow-twitch<span> muscles help enable long-endurance feats such as distance running.
</span>fast-twitch<span> muscles fatigue </span>faster<span> but are used in powerful bursts of movements like sprinting.</span>
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3 years ago
De donde eres responde y te doy corona
irakobra [83]

soy de texas, united states

8 0
3 years ago
An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma
Leya [2.2K]
Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
F=qvB \sin \theta
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
\theta the angle between the directions of v and B.

Re-arranging the formula, we find:
v= \frac{F}{qB \sin \theta}
and by substituting the data of the problem (the charge of the electron is q=1.6 \cdot 10^{-19} C), we find the velocity of the electron:
v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s
4 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0
3 years ago
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