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motikmotik
3 years ago
11

A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86

times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)

1. Yes
2. No
Physics
1 answer:
Vladimir [108]3 years ago
3 0

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

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The orbital radius of an electron in a hydrogen atom is 0.846 nm. What is its de Broglie wavelength?
kotykmax [81]

Answer:

The  value  is  \lambda   = 1.329 *10^{-9} \  m

Explanation:

From the question we are told that

  The  orbital radius is  r =  0.846nm =  0.846 *10^{-9} \ m

Generally the de Broglie wavelength is mathematically represented as

      \lambda  =  \frac{2 *  \pi  r}{4}

substituting values

     \lambda  =  \frac{ 2 * 3.142  *  0.846 *10^{-9}}{4}

    \lambda   = 1.329 *10^{-9} \  m

6 0
3 years ago
A cube of water 10 cm on a side is placed in a microwave beam having Ea = 11 kV/m‘ The microwaves illuminate one faceofthe cube,
Shtirlitz [24]

Answer:

The time is 133.5 sec.

Explanation:

Given that,

One side of cube = 10 cm

Intensity of electric field = 11 kV/m

Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.

We need to calculate the rate of energy transfer from the beam to the cube

Using formula of rate of energy

P=(0.80)IA

P=0.80\times\dfrac{c\mu_{0}E^2}{2}\times A

Put the value into the formula

P=0.80\times\dfrac{3\times10^{8}\times8.85\times10^{-12}\times(1.1\times10^{4})^2}{2}\times(10\times10^{-2})^2

P=1285.02\ W

We need to calculate the amount of heat

Using formula of heat

E =mc\Delta T

E =\rho Vc\Delta T

Put the value into the formula

E=1000\times(0.10)^3\times4186\times41

E=171626\ J

We need to calculate the time

Using formula of time

t=\dfrac{E}{P}

Put the value into the formula

t=\dfrac{171626}{1285.02}

t=133.5\ sec

Hence, The time is 133.5 sec.

3 0
3 years ago
Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in t
Over [174]
What are the choices ? 

Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation.  A few of them might be . . .

-- Kevin will have no trouble getting back in time for dinner.

-- Kevin will have no time to enjoy the scenery along the way.

-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.

           -- Speed = (distance covered) / (time to cover the distance) .

If time to cover the distance is zero, then speed is huge (infinite).

           -- Kinetic energy = (1/2) (mass) (speed)² .

If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.

         -- Mass = (Kevin's rest-mass) / √(1 - v²/c²)

-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
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8 0
3 years ago
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Diffusion occurs faster in gases than in liquids because _____.
jolli1 [7]
The correct answer to go in the blank would be A) The particles are moving faster.
5 0
3 years ago
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A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2
xz_007 [3.2K]

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

5 0
3 years ago
Read 2 more answers
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