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kondaur [170]
3 years ago
10

What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.20 nm?

Physics
1 answer:
Nat2105 [25]3 years ago
8 0
The momentum of a photon is given by the following relationship:
p= \frac{h}{\lambda}
where
h is the Planck constant
\lambda is the photon wavelength

For the photon in our problem, \lambda=0.20 nm = 0.20 \cdot 10^{-9}m, so its momentum is
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{0.20 \cdot 10^{-9} m}=3.3 \cdot 10^{-24} kg m/s

The electron must have the same momentum of this photon, and its momentum is given by (in the non-relativistic approximation)
p=mv
where
m is the electron mass
v is its speed

Re-arranging this formula, we can calculate the electron speed:
v= \frac{p}{m}= \frac{3.3 \cdot 10^{-24} kg m/s}{9.1 \cdot 10^{-31} kg} =3.6 \cdot 10^6 m/s

And this velocity is quite small compared to the speed of light (c=3 \cdot 10^8 m/s), so the non-relativistic approximation that we used for the electron's momentum is valid.
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The annoying sound from a mosquito is produced when it beats its wings at the average rate of 600. wingbeats per second. What is
jok3333 [9.3K]
The frequency is exactly the rate at which the bug beats its wings.
If that "600" that you mentioned is 600 beats per second, then
THAT's the frequency . . . 600 per second. (600 Hz) 
3 0
3 years ago
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

\Sigma F = F_{E}-W = 0 (Eq. 1)

Where:

F_{E} - Electrostatic force exerted on human, measured in Newton.

W - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

q\cdot E-m\cdot g = 0

q = \frac{m\cdot g}{E} (Eq. 2)

E - Electric field, measured in Newtons per Coloumb.

m - Mass, measured in kilograms.

g - Gravity acceleration, measured in meters per square second.

q - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that m = 60\,kg, g = 9.807\,\frac{m}{s^{2}} and E = -150\,\frac{N}{C}, the charge that a 60-kg human must have to overcome weight is:

q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }

q = -3.923\,C

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

F = \kappa \cdot \frac{q^{2}}{r^{2}} (Eq. 3)

Where:

\kappa - Electrostatic constant, measured in Newton-square meter per square Coulomb.

q - Electric charge, measured in Coulomb.

r - Distance between two people, measured in meters.

If we know that \kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q = -3.923\,C and r = 100\,m, then the force of repulsion between two people is:

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]

F = 13.851\times 10^{6}\,N

The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0
3 years ago
A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevatio
egoroff_w [7]

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

dtan13^0 = 0.231d

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

dtan4^0 = 0.07d

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

7 0
2 years ago
The amount of fluid displaced by a submerged object depends on its
Daniel [21]
The amount of fluid displaced by a submerged object depends on its volume.
7 0
3 years ago
Read 2 more answers
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
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