Angle, θ2 at which the light leaves mirror 2 is 56°
<u>Explanation:</u>
Given-
θ1 = 64°
So, α will also be 64°
According to the figure:
α + β = 90°
So,
β = 90° - α
= 90° - 64°
= 26°
β + γ + 120° = 180°
γ = 180° - 120° - β
γ = 180° - 120° - 26°
γ = 34°
γ + δ = 90°
δ = 90° - γ
δ = 90° - 34°
δ = 56°
According to the law of reflection,
angle of incidence = angle of reflection
θ2 = δ = 56°
Therefore, angle θ2 at which the light leaves mirror 2 is 56°
Answer:
3.88 * 10^(-15) J
Explanation:
We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.
First, we get the potential and potential energy:
Electric potential = E * r
E = electric field
r = distance between plates
Potential = 2.2 * 10^6 * 0.011
= 2.42 * 10^4 V
The relationship between electric potential and potential energy is:
P. E. = q*V
q = charge of electron = 1.602 * 10^(-19) C
P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)
P. E. = 3.88 * 10^(-15) J
No.
Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.
<h3>What is a random error?</h3>
Random error is defined as the deviation of the total error from its mean value due to chance.
Random errors can result from the instrument not being precise or from mistakes by the researcher.
Random errors can be minimized by taking multiple readings and averaging the results.
Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.
Learn more about random errors at: brainly.com/question/22041172
There is more thermal energy in the lake because there is more water which is more thermal energy
Answer:
The change in velocity is 15.83 [m/s]
Explanation:
Using the Newton's second law we have:
ΣF = m*a
The force in the graph is 185 N, therefore:
![185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]](https://tex.z-dn.net/?f=185%3D0.369%2Aa%5C%5CWhere%5C%5Ca%3Dacceleration%20made%20it%20by%20the%20force%20%5Bm%2Fs%5E2%5D)
![a=501.35[m/s^2]](https://tex.z-dn.net/?f=a%3D501.35%5Bm%2Fs%5E2%5D)
Now using the following kinematic equation:
![V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\](https://tex.z-dn.net/?f=V%5E%7B2%7D%3DVi%5E%7B2%7D%20%2B%202%2Aa%2A%28x-xi%29%20%5C%5Cwhere%5C%5CV%3Dfinal%20velocity%20%5Bm%2Fs%5D%5C%5CVi%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D%200%20the%20hockey%20disk%20is%20in%20rest%20when%20receives%20the%20hit.%5C%5C%20x%20%3D%20Final%20position%20%5Bm%5D%20%3D%200.4%20m%5C%5Cxi%20%3D%20initial%20position%20%5Bm%5D%20%3D%200.15m%5C%5C)
Now replacing the values:
![V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]](https://tex.z-dn.net/?f=V%5E%7B2%7D%3D0%20%2B%202%2A501.35%2A%280.4-0.15%29%5C%5C%20%5C%5CV%3D%2015.83%5Bm%2Fs%5D)
