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:
1.5024
Explanation:
Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.
R = 0.26 + 0.26 + 12 = 12.52
The bulb has a voltage of 2.88 volts across it. You can get the current from that.
i = E / R
i = 2.88 / 12 =
i = 0.24 amps.
Now you can get the voltage drop across the two cells.
E = ?
R = 0.26
i = 0.24 amps
E = 0.26 * 0.24
E = 0. 0624
Finally divide the 2.88 by 2 to get 1.44
Each cell has an emf of 1.44 + 0.0624 = 1.5024
If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;

- Distance of support from the left end;

- First mass;

- Distance of beam from the left end( m₁ is attached to );

- Second mass;

- Distance of beam from the right of the support( m₂ is attached to );

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, 
we divide both sides by 

Next, we make
, the subject of the formula
![x_1 = x - [ \frac{m_2x_2}{m_1} ]](https://tex.z-dn.net/?f=x_1%20%3D%20x%20-%20%5B%20%5Cfrac%7Bm_2x_2%7D%7Bm_1%7D%20%5D)
We substitute in our given values
![x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%20%5B%20%5Cfrac%7B61.7kg%5C%20%2A%20%5C%200.273m%7D%7B31.3kg%7D%20%5D)


Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Learn more; brainly.com/question/3882839
If you cannot get a chair to move across the floor, it is because static friction opposes your push. When you say static or kinetic friction the two object that facing each other are opposing each other. That's why you're having a hard time pushing the chair.