Average speed can be represented by the mathematical expression

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

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 gram

Bogdan [553]

Complete Question

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?

A His results are accurate but not precise.

B His results are neither accurate nor precise.

C His results are both accurate and precise

D His results are precise but not accurate.

Answer:

Correct option is A

Explanation:

From the question we are told that

The yield of the first test $k = 5.2 \ g$

The yield of the second is $u = 1.3 \ g$

The third yield is $p = 8.5 \ g$

The average yield $A = 5.0 \ g$

The know yield is $A_S = 5.1 \ g$

From the data given we see that

$A_S \ne A$

Since his average yield is closer to the known yield then the answer is accurate

But since the yield for each test are not repeated the answer is not precise

So the answer is accurate but not precise

4 0

3 years ago

A 7.25 kg7.25 kg block is sent up a ramp inclined at an angle ????=28.5°θ=28.5° from the horizontal. It is given an initial velo

Free_Kalibri [48]

Answer:

15.03 m

Explanation:

Given:

mass of the block, m = 7.25 kg

Angle, Θ = 28.5°

Initial speed of the block, v₀ = 15 m/s

let the distance traveled by the block be 's'

Now, applying the work energy theorem,

we have

$(m\times g\times\sin(\theta)\times s) + \mu_k\times mg\times s\times cos(\theta) = \frac{1}{2}\times m\times v^2$

on substituting the values in the above equation, we get

$(7.25\times 9.8\times\sin(28.5^o)\times s) + 0.326\times 7.25\times9.8\times s\times cos(28.5^o) = \frac{1}{2}\times 7.25\times 15^2$

or

$33.902\times s) +20.35\times s = 815.625$

or

$54.252\times s = 815.625$

s = 15.03 m

Hence, the block will travel 15.03 m up the ramp

6 0

3 years ago

How far does the moon move away from earth every year?

Oxana [17]

The moon is moving away from Earth at a rate of approximately 3.78 cm per year.
This migration of the Moon from the Earth is mainly due to the action of the Earth tides. It can be explained as follows:
- the Moon exerts a gravitational force on the Earth, which is stronger at the Equator (since the Equator is closer to the Moon), creating the tides
- However, the Earth rotates faster on its axis (one rotation every 24 hours) than the Moon (one rotation every 27 days), therefore the tidal bulge on Earth tries to pull the Moon "ahead" in its orbit. As a result, the Moon tends to sped up.
- As opposite reaction, the Earth tends to slow down in its rotation, with a loss of angular momentum. Since the angular momentum must be conserved, the radius of the orbit of the Moon becomes larger, and this explains why the Moon is moving away from the Earth.

4 0

4 years ago

Read 2 more answers

How does pressure affect surface tension

tatyana61 [14]

the effect of pressure on surface tension can be attributed in part to absorption of gas at the surface of the liquid and in part to an intrinsic decrease in density of the liquid in the neighborhood of the surface.

In the case of liquids , Owing to contact forces between the edge of the surface and the vessel, the surface acquires a curvature, and if the liquid rises up at the edges where it meets the vessel, the pressure will be less in the liquid than in the air, for points just below and just above the surface. The vessel exerts an upward force on the liquid. This is simply a matter of looking at the directions of forces acting, knowing that the surface is under tension.

3 0

3 years ago