A box is 30cm wide , 50cm long and 20cm high Calculate :

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.

Thus,

r = $\frac{mv}{qB}$ ÷ sinΘ

But, sinΘ = sin $90^{0}$ = 1.

So that;

r = $\frac{mv}{qB}$

= (9.11 × $10^{-31}$ × 121) ÷ (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ )

= 1.10231 × $10^{-28}$ ÷ 2.608 × $10^{-24}$

= 4.2266 × $10^{-5}$

= 4.23 × $10^{-5}$ m

The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

B. The frequency 'f' of the motion is called cyclotron frequency;

f = $\frac{qB}{2\pi m}$

= (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ ) ÷ (2 × $\frac{22}{7}$ × 9.11 × $10^{-31}$ )

= 2.608 × $10^{-24}$ ÷ 5.7263 × $10^{-30}$

= 455442.4323

f = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0

4 years ago

Read 2 more answers

A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet

ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S $_{y}$ = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326 E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S $_{y}$ = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N and S $_{y}$ > σ

3 0

3 years ago

The temperature -273°C is —.

ArbitrLikvidat [17]

Answer:

-273°C = Absolute zero

Explanation:

Average Kinetic energy of particles in matter refers to the temperature of matter. This value is not arbitrary zero temperature. It explains the lowest kinetic energy of the particles in matter.

4 0

4 years ago

which of the following laboratory tool would be most appropiate for measuring the approximate volume of a liquid?

SIZIF [17.4K]

Beaker would be most appropriate for measuring the approximate volume of a liquid.

7 0

3 years ago

Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.

Alexus [3.1K]

Answer:

a) 9.00 m b) 6.00 m/s c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = $\frac{1}{2}*a*t^{2}$ = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

$a = \frac{2*3.00m}{1s2}$ = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + $\frac{1}{2}*a*t^{2}$ = 6.00m/s*1s + $\frac{1}{2}*6.00 m/s2*1s^{2} = 9.00 m$

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s

3 0

3 years ago