Identify the following joint

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

dem82 [27]

Answer:

$r_\alpha=16cm$

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

$r=\frac{mv}{qB}(1)$

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

$K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\$

Solving this for mv:

$mv=\sqrt{2mK}(2)$

Replacing (2) in (1):

$r=\frac{\sqrt{2mK}}{qB}$

For protons, we have:

$r_p=\frac{\sqrt{2m_pK}}{eB}(3)$

For alpha particles, we have:

$r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)$

Dividing (4) in (3):

$\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm$

4 0

4 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip directions are at angles of 43.1° a

Bond [772]

Answer:

resolve shear stress = 22 MPa

Explanation:

Given data

slip plane α = 43.1°

slip directions β = 47.9°

shear stress = 20.7 MPa (3,000 psi)

applied stress =45 mPa (6,500 psi)

to find out

what stress will be necessary

solution

we know that

resolve shear stress = aplied stress × cosα × cosβ

resolve shear stress = 45 × cos(43.1) × cos(47.9)

resolve shear stress = 22 MPa

we can say that here single cristal will be yield

because resolve shear stress is bigger than critical shear stress

3 0

4 years ago

PLEASE HELP!!!!!The speakers in a sports stadium are 89.5 m from a fans seat. How much time does it take sound to travel from th

posledela

Answer:

t = space ÷ speed

t=89,5m ÷ 343 m/s=0,260932945s

7 0

3 years ago

Solar cells change into electricity.

ValentinkaMS [17]

Answer:

yes that is true because if you put the solar thing in the sunlight it gets converted to electricity

3 0

3 years ago

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A circuit consists of four 100W lamps

Ksivusya [100]

Complete question is;

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

150.42 V

Explanation:

We are told that the circuit consists of four 100W lamps.

We know that Power is given by the equation;

P = V²/R

Thus;

R = V²/P

Now, we are told that the four lamps are connected in parallel across a 230V supply.

Thus, V = 230 V

So resistance, R = 230²/100

R = 529 Ω

We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.

Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω

So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω

Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω

Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω

Thus;

circuit current = 230/2293.5 = 0.100283 A

Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V

7 0

4 years ago

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