Answer:
v = 2.94 m/s
Explanation:
When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.
Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.
Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means
(1/2)kx^2 = (1/2)mv^2
kx^2 = mv^2
v^2 = (kx^2)/m
v = sqrt((kx^2)/m)
v = x * sqrt(k/m)
v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg
v = 2.94 m/s
<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"
So, option B is your answer
Hope this helps!
</span>
Answer:
C
Explanation:
Vector A points up
Vector B points right
The combination must be both up and right which is C
Answer:
a
The number of fringe is z = 3 fringes
b
The ratio is 
Explanation:
a
From the question we are told that
The wavelength is 
The distance between the slit is 
The width of the slit is 
let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as
Substituting values
z = 3 fringes
b
From the question we are told that the order of the bright fringe is n = 3
Generally the intensity of a pattern is mathematically represented as
![I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%5B%5Cfrac%7B%5Cpi%20d%20sin%20%5Ctheta%7D%7B%5Clambda%7D%20%5D%5B%5Cfrac%7Bsin%20%28%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%20%7D%20%29%7D%7B%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%7D%20%7D%20%5D)
Where 
is the intensity of the central fringe
And Generally 
![I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20co%5E2%20%5B%20%5Cfrac%7B%5Cpi%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%5D%20%5B%5Cfrac%7B%5Cfrac%7Bsin%20%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%7D%7B%5Cfrac%7B%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%28n%20%5Cpi%29%5B%5Cfrac%7B%5Cfrac%7Bsin%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%29%7D%7B%20%5Cfrac%7B%20%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%20%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%283%20%5Cpi%29%20%5B%5Cfrac%7Bsin%20%28%5Cfrac%7B3%20%5Cpi%20%7D%7B6%7D%20%29%7D%7B%5Cfrac%7B3%20%5Cpi%7D%7B6%7D%20%7D%20%5D)
Answer:
36 kg
Explanation:
To answer this question, a few assumptions have to be made:
- That the temperature on the day is 35 °C
- That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant
Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J
Q = mcΔT
9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g
m = 36 000 g = 36 kg
