Answer:
Hope it may help you have a great day at school bye bye. Plz put it in brainliest answer my only request plz plz plz plz plz
Answer:
D
Explanation:
not sure but the answer is on quizlet . & i got a 100 so yeah
Answer:
Explanation:
angular frequency of resulting SHM ω = √(k/m
ω =√(100x32/1
ω = 56.57
Period T
ω = 2π /T = 56.57
T = 2π / 56.57
= .11 s .
spring is stretched by 1 inch so it will become the amplitude of frequency
A = 1 inch
when t = 0 , x ( displacement ) = A = 1 inch
equation of motion
x ( t ) in inch = A cosω t
= 1 cos 56.57t
x ( t ) in inch = 1 cos 56.57 t .
Answer:
the correct answer is 3 meters
Answer:
If the temperature of the solar surface is 5800 K then the approximate temperature of the sunspot is a) 4400 K.
Explanation:
The most straightforward way to solve this is using Stefan-Boltzmann law that states that I the energy radiated per unit surface area per unit time (watt per unit area
) of a black body is proportional to the fourth power of the temperature T of the body:

with
being the Stefan constant.
A black body is an idealized physical body that is a perfect absorber because it absorbs all incident electromagnetic radiation and is also an ideal emitter. The Sun is considered to be a black body at different layers and different temperatures.
We are told that the intensity of a sunspot
is found to be 3 times smaller than the intensity emitted by the solar surface
, that means that:

then using the expression of Stefan-Boltzmann law we get that

we cross out
and use the fourth root in each side of the equation
![\sqrt[4]{T_{sunspot} ^{4}}=\frac{\sqrt[4]{T_{surface} ^{4}}}{\sqrt[4]{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7BT_%7Bsunspot%7D%20%5E%7B4%7D%7D%3D%5Cfrac%7B%5Csqrt%5B4%5D%7BT_%7Bsurface%7D%20%5E%7B4%7D%7D%7D%7B%5Csqrt%5B4%5D%7B3%7D%7D)
![T_{sunspot}=\frac{T_{surface}}{\sqrt[4]{3} }](https://tex.z-dn.net/?f=T_%7Bsunspot%7D%3D%5Cfrac%7BT_%7Bsurface%7D%7D%7B%5Csqrt%5B4%5D%7B3%7D%20%7D)
then we use that


So finally we get that
