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fgiga [73]
3 years ago
14

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide

magnetic field isperpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is 700V.
Part A:What is the electron's speed?(Express your answer to two significant figures and include the appropriate units.)

Part B: If the potential difference between the plates is set to zero, what is the electron's radius of curvature in the magnetic field?(Express your answer to two significant figures and include the appropriate units.)
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0

Answer:

(a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2\ cm

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

v=\sqrt{\dfrac{2eV}{m}}

Put the value into the formula

v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}

v=15689290.81\ m/s

v=1.56\times10^{7}\ m/s

(b). We need to calculate the radius of electron

Using formula of centripetal force

\dfrac{mv^2}{r}=qvB

r=\dfrac{mv}{qB}

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}

r=0.042\ m

r=4.2\ cm

Hence, (a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2 cm

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You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

u =6.4013m/s

The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

h = 2.0906m

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

h = 2.0906+1.87

h = 3.9606m

4 0
4 years ago
A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067
Vesna [10]

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

P = mv

m = \dfrac{P}{v}

Where, P = momentum

v = speed

Put the value into the formula

m = \dfrac{2.8}{187}

m = 0.015\ kg

m = 15\ g

Hence, The mass of the bullet is 15 g.

4 0
3 years ago
A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land
erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + \frac{1}{2}at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = \sqrt{\frac{2h}{g} }.

⇒ t = \sqrt{\frac{2(1.3)}{9.81} } = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = \frac{s}{t}.

so, u = \frac{0.75}{0.51} = 1.45 m/s .

6 0
3 years ago
A rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Which statement
Step2247 [10]

Answer:

d. Its magnitude and its direction both remained the same.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

This ultimately implies that, the law of conservation of momentum states that if objects exert forces only on each other, their total momentum is conserved.

In this scenario, a rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Thus, the statement which correctly describes the momentum of the rubber ball is that its magnitude and its direction both remained the same because its velocity didn't change while returning to the thrower.

4 0
3 years ago
What is the net work done on the 20kg block while it moves the 4 meters?
Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

     Weight = 20 x 9.81

     Weight = 196.2 N

2.- Calculate the work done

     Work = 196.2 x 4

     Work = 784.8 J

5 0
4 years ago
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