Calculate the wavelength of a wave if 5 complete waves occupy a length of 20m

Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?

Lelu [443]

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

5 0

2 years ago

2) If the current in any circuit reach to infinity then its resistance becomes

In-s [12.5K]

Answer:

Zero according to ohm's law R= V / i

4 0

3 years ago

Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a fo

Dafna1 [17]

Answer: 75 N to the right

Explanation:

5 0

2 years ago

Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity i

AlladinOne [14]

Answer:

Explanation:

Atmospheric pressure = 7 x 10⁴ Pa

force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area

= 7 x 10⁴ x 3.14 x 2 x 2

= 87.92 x 10⁴ N

b )

weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m

Pressure x area

height x density x acceleration of gravity x π r²

= 10 x 415 x 6.2 x 3.14 x 2 x 2

=323168.8 N

c ) Pressure at a depth of 10 m

atmospheric pressure + pressure due to liquid column

= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)

= 7 x 10⁴ + 10 x 415 x 6.2

(7 + 2.57 )x 10⁴ Pa

9.57 x 10⁴ Pa

8 0

3 years ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago