Answer:
Core
Radiative zone
Convective zone
Photosphere
Chromosphere
Transient region
Corona
Ranks of layers based on their distance from the sun’s center
1st-corona
2nd-Transient region
3rd-chromosphere
4th-Photosphere
5th-convective zone
6th-radiative zone
7th-core
-- The speed of light in air is very close to 3 x 10⁸ m/s.
Whatever the actual number is, it's equivalent to roughly
7 times around the Earth in 1 second. So for this kind of
problem, you can assume that we see things at the same time
that they happen; don't bother worrying about how long it takes
for the light to reach you.
-- For sound, it's a different story. Sound in air only travels at
about 340 m/s. It takes sound almost 5 seconds to go 1 mile.
-- Now, the lightning and thunder happen at the same time.
The light travels to you at the speed of light, so you see the
lightning pretty much when it happens. But the sound of the
thunder comes poking along at 340 m/s, and arrives AFTER
the sight of the lightning.
The length of time between the sight and the sound is about
99.9999% the result of the time it takes the sound to reach you.
If the thunder arrived at you 3 seconds after the light did, then
the sound traveled
(340 m/s) x (3 s) = 1,020 meters .
(about 0.63 of a mile)
(If you're worried about ignoring the time it takes
for the light to reach you ...
It takes light 0.0000034 second to cover the same 1,020 meters,
so including it in the calculation would not change the answer.)
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
99 V
Explanation:
The effective voltage of an AC current (also called rms voltage) is given by
where
Vrms is the rms voltage
V0 is the peak voltage
In this problem, we know the effective voltage:
Therefore, we can re-arrange the equation to find the peak voltage:
Answer:
I started answering this before everything was erases, but briefly
W = 1/2 K x^2 so K = 2 W / x^2 = 6 * 25 = 150
Also F = - K X = 150 * 1/5 = 30 N at max extension
m = F / a = 30 /15 = 2 kg
Since max extension occurs at t = 0 you can write
x = A sin ω t
v = ω A cos ω t
a = -ω^2 A sin ω t note that there is no phase angle here
Also at .1 compression
.1 = .2 sin ω t and sin ω t = 1/2 and ω t = 30 deg = pi / 6
Substiture these values above to get velocity and acceleration when
ω t = 30 deg
