The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>
The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,
Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
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Answer:
Explanation:
What problem says can be written mathematically as:
Where:
The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:
Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:
Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:
Answer:
A.) 3605.6 N
B.) 33.7 degree
Explanation:
To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components
Resolving into x component :
Sum of forces = 3500 - 500 = 3000N
Resolving into y component:
Sum of forces = 2000N
Resultant force Fr = sqrt ( Fx^2 + Fy^2)
Fr = sqrt ( 3000^2 + 2000^2 )
Fr = sqrt ( 9000000 + 4000000 )
Fr = sqrt ( 13000000)
Fr = 3605.6 N
Therefore, resultant force acting on the wing is 3605.6 N
The direction of the vector will be:
Tan Ø = Fy / Fx
Substitute Fx and Fy into the formula
Tan Ø = 2000 / 3000
Tan Ø = 0.66666
Ø = tan^-1(0. 66666)
Ø = 33.7 degree.
Answer:
658.16N
Explanation:
Step one:
given data
mass m= 235kg
Force F= 760N
angle= 30 degrees
Required
The horizontal component of the force
Step two:
The horizontal component of the force
Fh= 760cos∅
Fh=760cos30
Fh=760*0.8660
Fh=658.16N
