Answer:
L = L0 (1 + c T) where c is the coefficient and T the change in temperature
L = 50 ( 1 + 2.05E-6 * 50) = 50.0051 cm
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
Answer:
12.7m/s
Explanation:
Given parameters:
Mass of the diver = 77kg
Height = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use one of the motion equations.
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + (2 x 9.8 x 8.18)
v² = 160.3
v = 12.7m/s
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,
c) threshold energy
h f =Ф
Explanation:
It's photoelectric effect was fully explained by Einstein by the expression
Knox = h f - fi
Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) wavelength is related to frequency
λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.
c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy
h f =Ф
