State two advantages of using an efficient water heater in a kitchen

How do buildings affect fog formation? Explain.<br> Please answer ASAP. I am marking brainliest!!

Kobotan [32]

Answer:

Buildings absorb heat during the day and radiate heat at night. So, temperatures at night become warmer than normal. This warmth prevents fog formation because fog formation requires low temperatures.

Explanation:

5 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1

Nesterboy [21]

Answer:

a) $a_c=3.41x10^{-5} \frac{m}{s^2}$

b) $a_c=3.34x10^{-5}\frac{m}{s^2}$

If we analyze the two values obtained for the centripetal acceleration we see that are similar. Based on this we can say that the centripetal force would be similar to the gravitational force between the Earth and Moon.

Explanation:

1) Notation and important concepts

Centripetal acceleration is defined as "The acceleration experienced while in uniform circular motion. It always points toward the center of rotation and is perpendicular to the linear velocity."

Angular frequency is defined as "(ω), or radial frequency, measures angular displacement per unit time and the units are usually degrees (or radians) per second. "

T = 27.3 d represent the time required by the Earth around an specific point given in the problem

G= universal constant $6.673x10^{-11}\frac{Nm^2}{kg^2}$

M= represent the mass of the Moon=7.35x10^{22}kg

2) Part a

$a=\frac{GM}{r^2}$ (1)

The Earth-Moon Distance Is $3.84x10^8 km$ (average Value) and both rotates at a point located about 4700 km from the center of Earth, so the radius for this case would be the difference between these two values

$r=(3.84x10^8 km)-(4.7x10^6)=3.793x10^8 m$

Since we have the radius now we can replace into equation (1)

$a=\frac{(6.673x10^{-11}\frac{Nm^2}{kg^2})(7.35x10^22 kg)}{(3.793x10^8 m)^2}=3.41x10^{-5} \frac{m}{s^2}$

And the acceleration due to the Moon's gravity would be $3.41x10^{-5} \frac{m}{s^2}$ at the point required.

3)Part b

For this case we can find the centripetal acceleration from this formula:

$a_c =r \omega^2$ (2)

But on this case we don't have the angular frequency so we can find it with this formula

$\omega =\frac{2\pi}{T}$ (3)

But since the period is on days we need to convert that into seconds

$27.3dx\frac{86400s}{1d}=2358720sec$

Replacing the value into equation (3) we got:

$\omega =\frac{2\pi}{2358720}=2.664x10^{-6}\frac{rad}{s}$

Now we can find the centripetal acceleration with the equation (2), the new radius on this case since our reference is the Earth and the point is located 4700km=4700000m from the center of Earth then the new value for the radius would be r=4700000m

$a_c =r \omega^2 =(4700000m)(2.664x10^{-6}\frac{rad}{s})^2=3.34x10^{-5}\frac{m}{s^2}$

If we analyze the two values obtained for the centripetal acceleration we see that are similar. Based on this we can say that the centripetal force would be similar to the gravitational force between the Earth and Moon.

5 0

3 years ago

If their were two dogs and 1 cat who would win hint: The Car

bixtya [17]

The car that hit the dog and cat would win

8 0

3 years ago

Read 2 more answers

List the characteristics properties of all waves. at what speed do electromagnetic waves travel in a vacuum

ad-work [718]

All electromagnetic waves travel at

299,792,458 meters per second

in vacuum.

7 0

4 years ago

Problem: A lossless 50-Ω transmission line is terminated in a load with ZL = (50 + j25) Ω. Use the Smith chart to find the follo

Nadya [2.5K]

Answer: (a). ΓL = 0.246 < 75°

(b). S = 1.7

(c). Zin = (30-j)λ

(d). jreal = Arc Po = 0.105λ

(e). jmax = jreal = 0.105λ

Explanation:

attached is a document to help in understanding.

So we will begin with a step by step analysis of the problem.

from the diagram we have that ZL = (50 + j25) Ω.

where ZL = ZL / Z₀ = 50 + j25 / 50 = 1 + j0.5

so we mark this on the chart as point 'P'

(a) ΓL = mP/m 'P' < Θ L = 1.7/6.9 < 75°

ΓL = 0.246 < 75°

(b) This s-circle 's' is given thus s = r = 1.7 on the RHS of the chart

S = 1.7

(c) we are to calculate the input impedance;

ζin = Q = 0.6 - j0.02

therefore Zin = Z₀ζin = 50(0.6 - j0.02) = (30-j)λ

Zin = (30-j)λ

(d) here we are taking R as the diameter opposite of Q on the s=circle

so R = γin = 1.7 + j0.02

yin = yo (γin) = (1.7+j0.02) / 50 = (34 + j0.4)ms

yin = (34 + j0.4)ms

(e) move from 'p' on s-circle to 'o'

where maximum impedance = Znxl = Zos

which gives jreal = Arc Po = 0.105λ

(f) jmax = jreal = 0.105λ

cheers i hope this helps

3 0

3 years ago

State two advantages of using an efficient water heater in a kitchen​

State two advantages of using an efficient water heater in a kitchen