The area of the Earth (Ae) that is irradiated by is given by:
Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2
On the Earth, insolation (We) = Psun/Ae
Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W
Answer:
1.93 m/s
Explanation:
Parameters given:
Mass = 4.5g = 0.0045kg
Spring constant = 8.0 N/m
Length of barrel = 13 cm = 0.013m
Frictional force = 0.035N
Compression = 5.8 cm = 0.058m
First, we find the P. E. stored in the spring:
P. E. = ½*k*x²
P. E. = ½ * 8 * 0.058² = 0.013J
Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:
Work = Force * distance
The distance here is the length of the barrel.
Work = 0.035 * 0.13 = 0.0046 J
The kinetic energy of the sphere can now be found:
K. E. = P. E. - Work done
K. E. = 0.013 - 0.0046 = 0.0084J
We can now find the speed using the formula for K. E.:
K. E. = ½*m*v²
0.0084 = ½ * 0.0045 * v²
v² = 0.0084/0.00255 = 3.733
=> v = 1.93 m/s
Answer:
For example, the toes are anterior to the heel, and the popliteus is posterior to the patella. Superior and inferior, which describe a position above (superior) or below (inferior) another part of the body. For example, the orbits are superior to the oris, and the pelvis is inferior to the abdomen.
Explanation:
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.
Answer: 
<u>Explanation:</u>
A linear equation is of the form: y = mx + b where
- m is the slope
- b is the y-intercept (where it crosses the y-axis)
x + 4y = 16
4y = -x + 16


The y-intercept (b) = 4
Next, find the slope given point (4, 5) and b = 4
