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andrey2020 [161]
3 years ago
14

What will most likely happen if a sound wave moves from the air through a solid?

Physics
2 answers:
stich3 [128]3 years ago
6 0

Answer:

It will increase in speed

Explanation:

kobusy [5.1K]3 years ago
5 0

Answer:

C. it will increase in speed

Explanation:

Edge 2021

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Comparison between copper properties and aluminium properties​
mixas84 [53]
Hopes this helps:

Answer: Aluminum has 61 percent of the conductivity of copper, but has only 30 percent of the weight of copper. That means that a bare wire of aluminum weights half as much as a bare wire of copper that has the same electrical resistance. Aluminum is generally more inexpensive when compared to copper conductors.
5 0
2 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob
pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

T^2 = \frac{4\pi^2}{GM}a^3

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s

PART A) Replacing the values to find a, we have

a^3= \frac{T^2 GM}{4\pi^2}

a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}

a^3 = 6.46632*10^{39}

a = 1.86303*10^{13}m

Therefore the semimajor axis is 1.86303*10^{13}m

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

R = a(1-e)

R = 1.86303*10^{13}(1-0.997)

R= 5.58*10^{10}m

7 0
3 years ago
Is N2 a triple bond
krek1111 [17]
Yes, N2 is a triple bond.
Hope it helps!
5 0
3 years ago
What is the velocity of the girl in Figure b? Please help me!!!!<br> Is the answer A,B,C,D,E, or F?
attashe74 [19]
There is a Blank screen

6 0
2 years ago
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