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3 years ago

6

A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, the final

particle energy is proportional to:

1 answer:

Archy [21]3 years ago

5 0

Answer:

the final particle energy is proportional to

$Emax \infty\ R^2$

Explanation:

The computation of the final particle energy is as follows;

As we know that

Energy is respect with the radious as

$Emax = \frac{q^2R^2B^2}{2m}$

In the case when other quantities would remains the same or constant

So, the final particle energy is proportional to

$Emax \infty\ R^2$

The same is to be considered and relevant too

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Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0

3 years ago

What net force would be required for a 50.5 kg wolf to accelerate at 5.0 m/s^2

Basile [38]

Answer:

$\boxed {\boxed {\sf 252.5 \ Newtons }}$

Explanation:

Force is the push or pull on object that can cause different things, like acceleration. According to Newton's 2nd Law of Motion, it is the product of mass and acceleration.

$F=m*a$

The mass of the wolf is 50.5 kilograms and the acceleration is 5.0 meters per square second. Therefore:

$m= 50.5 \ kg$

$a= 5.0 \ m/s^2$

Substitute the values into the formula.

$F= 50.5 \ kg * 5.0 \ m/s^2$

Multiply.

$F=252.5 \ kg*m/s^2$

1 kilogram meter per square second is equal to 1 Newton.
Our answer of 252.5 kg*m/s² is equal to 252.5 Newtons.

$F= 252.5 \ N$

The force required is <u>252.5 Newtons</u>.

4 0

3 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

b

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

morpeh [17]

Velocity is not changing.

Acceleration is zero.

Net force is zero.

4 0

3 years ago

Read 2 more answers

What is the power of a kitchen blender if it can perform 2000 joules of work in 5 seconds ?

soldier1979 [14.2K]

Power = (work done) / (time to do the work)

Power = (2000 J) / (5 sec)

<em>Power = 400 watts</em>

7 0

3 years ago