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Step2247 [10]
3 years ago
5

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

tarting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 102 N/C and separation between the charged plates is 2.0 cm. Determine the horizontal distance travelled by the electron when it hits the plate.
Physics
1 answer:
max2010maxim [7]3 years ago
4 0

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

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Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis
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Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{Mm}{r^{2}} (1)

Where:

F is the gravitational force between Earth and Moon

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M=5.972(10)^{24} kg is the mass of the Earth

m=7.349(10)^{22} kg is the mass of the Moon

r=3.9(10)^{8} m is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to F:

F=m.a_{C} (2)

Where a_{C} is the centripetal acceleration given by:

a_{C}=\frac{V^{2}}{r} (3)  

Being V the orbital velocity of the moon

Making (1)=(2):

m.a_{C}=G\frac{Mm}{r^{2}} (4)

Simplifying:

a_{C}=G\frac{M}{r^{2}} (5)

Making (5)=(3):

\frac{V^{2}}{r}=G\frac{M}{r^{2}} (6)  

Finding V:

V=\sqrt{\frac{GM}{r}} (7)

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}} (8)

Finally:

V=1010.92 m/s

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8 0
2 years ago
An object is located 50.0 cm from a concave mirror. The magnitude of the mirror focal length is 25.0 cm. What is the image dista
ioda

Answer:

Correct answer: C. 50 cm

Explanation:

Given data:

The distance of the object from the top of the concave mirror o = 50.0 cm

The magnitude of the concave mirror focal length 25.0 cm.

Required : Image distance d = ?

If we know the focal length we can calculate the center of the curve of the mirror

r = 2 · f = 2 · 25 = 50 cm

If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.

We conclude that the image distance is 50 cm.

We will now prove this using the formula:

1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50

1/d = 1/50 => d = 50 cm

God is with you!!!

6 0
3 years ago
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