Answer:
Approximately
.
Explanation:
Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.
Look up the Rydberg constant for hydrogen:
.
Look up the speed of light in vacuum:
.
Look up Planck's constant:
.
Apply the Rydberg formula to find the wavelength
(in vacuum) of the photon in question:
.
The frequency of that photon would be:
.
Combine this expression with the Rydberg formula to find the frequency of this photon:
.
Apply the Einstein-Planck equation to find the energy of this photon:
.
(Rounded to three significant figures.)
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
<u>W = 12.96 J</u>
The less valence electrons the more reactive the element is.
Answer:
V = P = 0 m/s
Explanation:
When a pendulum bob is given an initial displacement or the initial velocity, it starts to execute periodic motion or simple harmonic motion. During this motion the kinetic and potential energy keeps interconverting. The kinetic energy becomes maximum at the lowest point, that is the mean point. Hence, the velocity is maximum at this point, as well. Similarly, at both extreme positions the potential energy becomes maximum due to maximum height, while the kinetic energy becomes zero at the highest point, that is extreme positions. At these, positions the velocity will be minimum and it will be zero due to zero kinetic energy. Hence, at both extreme positions the bob stops momentarily before, reversing the direction. Hence,
<u>V = P = 0 m/s</u>
Answer:
Explanation:
Component of force F₁ in right direction = F₁cos37
= 12 x cos37 = 9.58 N .
Component of force F₂ in vertically upward direction = F₁sin37
= 12 x sin37 = 7.22 N .
a ) Let normal force be R
R + F₁sin37 = mg
R + 7.22 = 1.5 x 9.8 = 14.7
R = 7.48 N .
b )
Net force in horizontal direction = F₂ - F₁cos37
= F₂ - 9.58
This is equal to zero as body is moving with zero acceleration
F₂ - 9.58 = 0
F₂ = 9.58 N
c ) If body is moving with acceleration of 2.5 m /s² along the direction of F₂
F₂ - 9.58 = 1.5 x 2.5 = 3.75
F₂ = 13.33 N .