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IceJOKER [234]
3 years ago
9

An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency doe

s the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)
Physics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

Frequency, f=1.2\times 10^{10}\ Hz

Explanation:

It is given that,

Magnetic field, B = 0.43 T

We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :

f=\dfrac{qB}{2\pi m}

f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}

f=1.2\times 10^{10}\ Hz

Hence, this is the required solution.

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Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?
Korolek [52]

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

8 0
3 years ago
List 3 cases in which potential energy becomes kinetic energy and three cases in which kinetic energy becomes potential energy.
Mrrafil [7]

Answer:

when you open a can of pop

when you jump on your bed

Explanation:

3 0
3 years ago
If u speed up from rest to 12 m/s in 3 seconds, what is your acceleration
irakobra [83]
4

Just divide 12 by 3, so if it takes 3 seconds, then every second, it goes up 4.
4 0
4 years ago
Read 2 more answers
If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat
Mamont248 [21]

<h2>The option ( c ) is correct </h2>

Explanation:

As the frequency of oscillation of any oscillator is doubled

The velocity of sound  v = νλ

here ν is the frequency and λ is the wavelength

Now if ν becomes double , the wavelength λ becomes one half . The velocity of sound remains the same in the same medium .

Thus option ( c ) is correct

7 0
3 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
3 years ago
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