The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

Answers

Answer 1

Answer: (a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= (1)/(2)mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ (2K)/(m) }= \sqrt{ (2 \cdot 2.20 \cdot 10^(-15) J)/(9.1 \cdot 10^(-31) kg) }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m (v^2)/(r)$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= (mv)/(qB)= ((9.1 \cdot 10^(-31) kg)(6.95 \cdot 10^7 m/s))/((1.6 \cdot 10^(-19) C)(3.00 \cdot 10^(-5) T))=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = (v^2)/(r)= ((6.95 \cdot 10^7 m/s)^2)/(13.18 m)=3.66 \cdot 10^(14) m/s^2$

Related Questions

On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before shooting over Niagara Falls. After moving over the edge, the water drops 53 mm to the water below.a. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom? b. Express your answer to two significant figures and include the appropriate units. c. How far does the water move horizontally during this time? d. Express your answer to two significant figures and include the appropriate units.
A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?
A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.
A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?
What is an inexpensive, portable, and common way to assess body fat in the fitness industry?DEXABioelectrical impedanceSkinfold testingHydrostatic weighing

A fish swims 10 cm from the front wall of an aquarium that is 35cm wide. The front wall of the aquarium is glass with negligible thickness, but the back wall is a plane mirror. A person looks through the front wall and watches both the fish and its reflection in the mirror.Part A:What is the apparent distance from the front wall of the aquarium to the fish?Part B: What is the apparent distance from the front wall of the aquarium to the image of the fish in the mirror?

Answers

El problema es un caso generalmente tipico en optica concerniente a Apparent depth vs real depth

We see the objects closer than their real depth to the surface. We see objects only if the rays coming from them reaches our eyes.

The equation is given by,

$D_a = (D_r)/(\eta)$

Where,

$D_a =$ Apparenth depth

$D_r =$ Real depth

$\eta =$ Refractive index of the medium of object

For water $\eta$ is equal to 1.33

I attach an image of the theory that could help clarify the measurements.

We have,

$D_a = (D_r)/(\eta)$

$D_a = (10)/(1.333)$

$D_a = 7.5cm$

Therefore the apparent distance between the front wall of the aquarium to the fish is 7.5cm

B) The distance between fish and mirror is given by,

$d=35-10= 25$

So we have that real distance from the front wall of to image of fish is

$dr=25+35=60cm$

Applying our equation we have that,

$D_a = (D_r)/(\eta)$

$D_a = (60)/(1.333)$

$D_a = 45.1cm$

Therefore the apparent distance from the front wall of the aquarium to the image of the fish is 45.1cm

Final answer:

The apparent distance from the front of the aquarium to the fish is 10 cm, and the apparent distance from the front of the aquarium to the image of the fish in the mirror is 35 cm.

Explanation:

Part A: When observing the fish in the aquarium, the apparent distance from the front wall of the aquarium to the fish is simply the actual distance. This is because the observation is being made directly through the glass which has negligible thickness. Therefore, the apparent distance to the fish is 10 cm.

Part B: The image of the fish in the mirror will seem farther away than the fish itself. This is due to the fact that light reflects off the mirror and travels the distance of the aquarium twice. Hence, the total distance traveled by the light is the distance to the fish plus the distance from the fish to the mirror which is 35 cm - 10 cm = 25 cm. Thus, the apparent distance from the front wall of the aquarium to the image of the fish in the mirror is 10 cm (to the fish) + 25 cm (to the mirror) = 35 cm.

Learn more about Apparent Distance and Refraction here:

brainly.com/question/32255407

#SPJ3

Two narrow slits separated by 0.30 mm are illuminated with light of wavelength 496 nm. (a) How far are the first three bright fringes from the center of the pattern if observed on the screen 130 cm distant? (b) How far are the first three dark fringes from the center of the pattern?

Answers

Answer:

Explanation:

d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m

λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m

n = order of the bright fringe

D = screen distance = 130 cm = 1.30 m

$x_(n)$ = Position of nth bright fringe

Position of nth bright fringe is given as

$x_(n) =( n D \lambda )/(d)$

For n = 1

$x_(1) =( (1) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(1) = 2.15* 10^(-3)m$

For n = 2

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 4.30* 10^(-3)m$

For n = 3

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 6.45* 10^(-3)m$

Position of nth dark fringe is given as

$y_(n) =( (2n+1) D \lambda )/(2d)$

For n = 1

$y_(1) =( (2(1)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(1) = 3.22* 10^(-3)m$

For n = 2

$y_(2) =( (2(2)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(2) = 5.4* 10^(-3)m$

For n = 3

$y_(3) =( (2(3)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$x_(3) = 7.5* 10^(-3)m$

g A particular guitar string has a length of 60.0 cm, and a mass per unit length of 2.00 grams/meter. You hear a pure tone of 660 Hz when a particular standing wave, represented by the animation above, is excited on the string. Calculate the wavelength of this standing wave.

Answers

Answer:

wavelength of the standing wave will be equal to 30 cm

Explanation:

We have given length of the guitar string L = 60 cm

Mass per unit length $\mu =2gram/m$

Frequency is given f = 660 Hz

We have to find the wavelength of the standing wave

Length of the string will be 2 times of the wavelength of the wave

So $L=2\lambda$

$\lambda =(L)/(2)=(60)/(2)=30cm$

So wavelength of the standing wave will be equal to 30 cm

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

Look at the figure below and calculate the length of side y.A. 8.5
B. 17
12
y
C. 6
O D. 12
45
Х

Answers

Answer:

I want to say a because you want to subtract and simplify

Which best describes a reference frame?

Answers

Answer:

C a position from which something is observed

om edu 2021

Explanation:

Answer: A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes.

Other Questions

Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?

Explain what happent to the pressure exerted by an object when the area over which it is exerted:a) increase b) decrease

Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Assume the electric field has a magnitude of 59.0 m V , and the particle has a net positive charge of Q=+5.40 C. Calculate the magnitude of the electric potential difference, \vert \Delta V \vert = \vert V_B - V_A \vert∣ΔV∣=∣V B −V A ∣.