The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J
For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
Answer:
0.6 m
Explanation:
When a spring is compressed it stores potential energy. This energy is:
Ep = 1/2 * k * x^2
Being x the distance it compressed/stretched.
When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.
The potential energy of the ice cube is:
Ep = m * g * h
This is vertical height and is related to the distance up the slope by:
sin(a) = h/d
h = sin(a) * d
Replacing:
Ep = m * g * sin(a) * d
Equating both potential energies:
1/2 * k * x^2 = m * g * sin(a) * d
d = (1/2 * k * x^2) / (m * g * sin(a))
d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.
4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>
Answer:
B. 17,705.1 J
Explanation:
The hear released when the mercury condenses into a liquid is given by:
where
m = 0.06 kg is the mass of the mercury
is the latent heat of vaporization
For mercury, the latent heat of vaporization is 
, so the heat released during the process is:
So, the closest option is
B. 17,705.1 J
Solution :
Given :
Rectangular wingspan
Length,L = 17.5 m
Chord, c = 3 m
Free stream velocity of flow, 
= 200 m/s
Given that the flow is laminar.
So boundary layer thickness,
= 0.0024 m
The dynamic pressure, 
The skin friction drag co-efficient is given by
= 0.00021
= 270 N
Therefore the net drag = 270 x 2
= 540 N
