If a photon has a frequency of 5.20 x 10^14 hertz, what is the energy of the photon ? Given : Planck's constant is 6.63 x 10^-34 joule-seconds.


Answer 1
Answer: For this you would use planck's equation.

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14  x  6.63 x 10^-34
= 3.45 x 10^-19 Joules

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A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocityof the stroller after it has traveled 6.32 m?


You're going to use the constant acceleration motion equation for velocity and displacement:

(V)final² = (V)initial²+2a(dx)



dx=6.32 m


(V)final= UNKNOWN

(V)final= 2.51396m/s

Anatomy of a Wave worksheet can someone help me out with the answers????


Part 1: here are the answers in order

A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?


Final answer:

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.


To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

Learn more about Projectile Motion here:



On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0ºC and completely melts to 0ºC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s) ?



P=13.5 W


In this case, power is the rate of transferring heat per unit time:

P=(Q)/(\Delta t)(1)

The heat is given by the formula of the latent heat of fusion, since the ice is melting.


Here m is the ice's mass and L_f is the heat of fusion of ice. Recall that one day has 86400 seconds. Replacing (2) in (1) and solving:

P=(mL_f)/(\Delta t)\nP=(3.5kg(334*10^3(J)/(kg)))/(86400s)\nP=13.5 W

Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.


The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

Learn about free-body diagram here brainly.com/question/10148657



4. It is in static equilibrium with a net force of 0 N.


Just got it right :)

For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .



Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²


Please see attachment below.


z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

Final answer:

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.


The question is about understanding kinematics in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two components, one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the Pythagorean theorem: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the acceleration is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².

Learn more about Motion in Cylindrical Coordinates here: