Answer:
from Late Latin convectionem
Explanation:
Answer:
c. Solar eclipses would be much more frequent.
Explanation:
The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.
A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.
The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.
So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>
So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.
the answer is: c.
Answer:
22,800 years
Explanation:
Half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.0625 = (½)^(t / 5700)
log 0.0625 = (t / 5700) log 0.5
4 = t / 5700
t = 22,800
It takes 22,800 years.
Hello!
We can use the following equation for calculating power dissipated by a resistor:
![P = i^2R](https://tex.z-dn.net/?f=P%20%3D%20i%5E2R)
P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)
Plug in the known values and solve.
![P = (2.0^2)(50) = \boxed{\text{ B. }200 W}](https://tex.z-dn.net/?f=P%20%3D%20%282.0%5E2%29%2850%29%20%3D%20%5Cboxed%7B%5Ctext%7B%20B.%20%7D200%20W%7D)
To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency
![ma = \text{Mass}\times \text{Acceleration}](https://tex.z-dn.net/?f=ma%20%3D%20%5Ctext%7BMass%7D%5Ctimes%20%5Ctext%7BAcceleration%7D)
![ma = kg \cdot \frac{m}{s^2}](https://tex.z-dn.net/?f=ma%20%3D%20kg%20%5Ccdot%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
At the same time we have that
![\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Ctext%7BMass%7D%5Ctimes%20%5Ctext%7BVelocity%7D%5E2)
![\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20kg%20%28%20%5Cfrac%7Bm%7D%7Bs%7D%29%5E2)
![\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20kg%20%5Ccdot%20%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D)
Therefore there is not have same units and both are not consistent and the correct answer is B.