Identify 3 preys and their preys

A cyclotron is designed to accelerate protons (mass 1.67 x 10^-27 kg) up to a kinetic energy of 2.5 x 10^-13 J. If the magnetic

Dafna11 [192]

Answer:

24 cm

Explanation:

Given:

Mass of proton = 1.67 × 10⁻²⁷ Kg

kinetic energy = 2.5 × 10⁻¹³ J

magnetic field in the cyclotron, B = 0.75 T

Now,

Kinetic energy = $\frac{1}{2}mv^2$ = 2.5 × 10⁻¹³ J

where, v is the velocity of the electron

or

$\frac{1}{2}\times1.67\times10^{-27}\times v^2$ = 2.5 × 10⁻¹³ J

or

v² = 2.99 × 10¹⁴

or

v = 1.73 × 10⁷ m/s

also,

centripetal force = magnetic force

or

$\frac{mv^2}{r}$ = qvB

q is the charge of the electron

r is the radius of the dipole magnets

on substituting the respective values, we get

$\frac{1.67\times10^{-27}\times1.73\times10^7}{r}$ = 1.6 × 10⁻¹⁹ × 0.75

or

r = 0.2408 m ≈ 24 cm

Hence, the correct answer is 24 cm

6 0

3 years ago

Why do astronauts muscles weaken?

Pavlova-9 [17]

They work on a weightless planet, therefore they have very little muscle control since they float in space !!

3 0

3 years ago

Calculate the Fermi energy and the conductivity at room temperature for germanium containing 5×10^16 arsenic atoms per cubic cen

Cloud [144]

Answer:

The Fermi energy is 0.568 eV and the conductivity at room temperature is 31.24 (Ωcm)⁻¹

Explanation:

Data given:

Nd = doping concentration = 5x10⁶/cm³

According the mass action law, the hole concentration is:

$P_{0} =\frac{n_{i}^{2} }{n_{0} } ,n_{i}=2x10^{13} /cm^{3}$

$E_{f} =E_{i} +KTln(\frac{n_{D} }{n_{i} } ),K=8.61x10^{-19} ,KT=0.026eV$

$E_{i} =\frac{E_{0} }{2} =\frac{0.63}{2} =0.315eV\\E_{f}=0.315+0.203=0.568eV$

The conductivity of n-type of semi-conductor is equal to:

The mobility of Germanium is 3900 cm²/Vs

σ = 1.6x10⁻¹⁹ * 5x10¹⁶ * 3900 = 31.24 (Ωcm)⁻¹

6 0

3 years ago

The process shown in this diagram contributed great amounts of heat to the young planet Earth and is best known as radioactive

Elis [28]

<span>The process shown in this diagram contributed great amounts of heat to the young planet Earth and is best known as radioactive decay. Decay is known to release large amounts of heat. </span>

4 0

3 years ago

Read 2 more answers

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t

statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

$\omega=\omega_{0}+\alpha t$

Put the value in the equation

$\omega=0.17+1.3\times1.7$

$\omega=2.38(k)\ m/s$

We need to calculate the angular displacement

Using angular equation of motion

$\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}$

Put the value in the equation

$\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}$

$\theta=2.1675\times\dfrac{180}{\pi}$

$\theta= 124.18^{\circ}$

We need to calculate the velocity at point A

Using equation of motion

$v_{A}=v_{0}+\omega\times r$

Put the value into the formula

$v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))$

$v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i$

$v_{A}=(-0.267j-0.393i)\ m/s$

We need to calculate the acceleration at point A

Using equation of motion

$a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)$

Put the value in the equation

$a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=-0.146j-0.215i−0.636i+0.937j$

$a_{A}=0.791j-0.851i$

$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

3 0

4 years ago