The acceleration due to gravity is equal to second squared. (round to 2 significant digits)

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

(c) ratio = 1800

Explanation: (a) Potential difference is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

so

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

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

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

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

5 0

4 years ago

I’m in the middle of a test right now. If anybody knows this can they help

babymother [125]

Answer:

C

Explanation:

momentum = mass ×velocity

A. mv = 200

B. mv = 300

C. mv= 400

D. mv= 200

highest is C.

6 0

3 years ago

A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops

Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are $P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg$ and $v_{2} =0.4v_{1}$ from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the $T_{sat}=179.88^{0} C \leq T_{1}$ from the table 1 we get: $v_{1} =0.30661(m^{3}/Kg)$ . Now we have second conditions as: $P_{2}=1(MPa), T_{2}=250^{0}C$ so from the same table we can see the state still superheated and we get $v_{2}=0.23275(m^{3}/Kg)$ , knowing that it's a isobaric process we can find the compression's work as: $W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)$ so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and: $W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)$ and the second process is isochoric: $W_{2}=zero$ ,now $W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)$ . Finally to get the temperarure at the final state in part (b) we get: $v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)$ from table 2 (see attached) we compare $v_{f}$ and $v_{g}$ at the saturated water table and find the following: $v_{f}=0.001093(m^{3}/Kg)$ , so we know that the final state phase is a satured mixture and we get the temperature at the final state as: $T_{2} =T_{sat} =151.83^{0}C$ .

3 0

4 years ago

The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi

iragen [17]

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

3 0

3 years ago

Which missing item would complete this beta decay reaction?

Flauer [41]

For us to understand the missing item that would complete beta decay reaction, we need to achieve in depth understanding of chemical formulas and nuclear symbols. Next is to have great comprehension of the following points:
1.) Neutron in nucleus breaks down and changes into a proton.
2) Then it emits an electron, as well as an anti-neutrino which go into space.
3) Lastly, atomic number continuously goes UP while mass number remains unchanged.

8 0

3 years ago