The third one sliding friction
Explanation:
Answer:
W = 100000 J = 100 KJ
Explanation:
Here we will use the most basic and general formula of work, which is as follows:

where,
W = Work Done = ?
F = Force Required = 200 N
d = Length of Track = 500 m
Therefore,

<u>W = 100000 J = 100 KJ</u>
The photoelectric effect is obtained when you shine a light on a material, resulting in the emission of electrons.
The kinetic energy of the electrons depends on the frequency of the light:
K = h(f - f₀)
where:
K = kinetic energy
h = Planck constant
f = light frequency
f₀ = threshold frequency
Rearranging the formula in the form y = m·x + b, we get:
K = hf - hf₀
where:
K = dependent variable
f = <span>indipendent variable
h = slope
hf</span>₀ = y-intercept
Every material has its own threshold frequency, therefore, what stays constant for all the materials is h = Planck constant (see picture attached).
Hence, the correct answer is
C) the slope.
Answer:
1/i + 1/o = 1/f thin lens equation
i = 33 * 8.9 / (33 - 8.9) = 12.2 cm to right of first lens
27 - 12.2 = 14.8 cm to left of second lens
i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens
Answer:
0.3405V
Explanation:
#Given a magnetic field of
, diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:

we can now calculate the induced emf,
:

Hence, the induced emf of the loop is 0.3405V