a beam of 1mev electrons strike a thick target. for a beam current of 100 microampere, find the power dissipated in the target

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Explanation: hope this helps ;)

Answer:

9,375

Explanation:

Data provided

The mass of the car m = 1500 Kg.

The diameter of the circular track D = 200 m.

For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-

The radius of the circular path is

[tex]R = \frac{D}{2}[/tex]

[tex]= \frac{200}{2}[/tex]

= 100 m

after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula

[tex]Force F = \frac{mv^2}{R}[/tex]

[tex]= \frac{1500\times (25)^2}{100}[/tex]

= 9,375

Therefore for computing the magnitude of the centripetal force we simply applied the above formula.

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

Answer:

  F ’= F 0.25

Explanation:

This problem refers to the electric force, which is described by Coulomb's law

        F = k q₁ q₂ / r²

where k is the Coulomb constant, q the charges and r the separation between them.

The initial conditions are

       F = k q_A q_B / d²

they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d

       F ’= k (q / 4 q / 4) / (0.5 d)²

       F ’= k q / 16 / 0.25 d²

       F ’= k q² / d²    0.0625 / 0.25

       F ’= F 0.25

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.

Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]

where,

If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:

[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Learn more about Coulomb's law here: https://brainly.com/question/506926

The  lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of  the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So,  The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.

Learn more about diffraction here:

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Answer:

6 N

Explanation:

From the laws of friction

F = ¶R = 0.3 × 20 = 6 N

The force of friction opposing the block's motion is 6 N.

The given parameters;

The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.

F = ma

Fₓ = μF

Substitute the given parameters to calculate the frictional force on the object.

Fₓ = 0.3 x 20

Fₓ = 6 N

Thus, the force of friction opposing the block's motion is 6 N.

Learn more here: https://brainly.com/question/18247518

The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).

So if a wave completes one cycle in 2 seconds, then that is its period.

Answer:

b. only ray that reflects back in the same direction it came from

Explanation:

C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.

Correct answer is option B.

C-ray is the only ray that reflects back in the same direction it came from.

Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.

Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>    [tex]e * V = \frac{1}{2} * m * v^2[/tex]

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

Answer:

[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]

Explanation:

[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]

[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]

[tex]\displaystyle \mathrm{a = 4.5}[/tex]

[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]

Answer:

  d sin tea = m λ

Explanation:

When we have a two-slit system, the optical path difference determines whether the intensity reaching an observation screen is maximum or zero.

To find this difference in optical path, we assume that the screen is much farther than the gap is, we draw a perpendicular from ray 1 to the second ray

       OP = d sin θ

now to have constructive interference and see a bright line this leg must be an integer number of wavelengths, ose

            d sin tea = m λ

where

d             is the distance between the two slits

θ     complexion the angle sea the point hold it between the two slits

λ    the wavelength of the coherent light used

m   an integer, which counts the number of lines of interference

Units in the SI system

d, lam in meters

θ degrees

m an integer

Answer:

It is the mean of 460 swimmers

Explanation:

In this question, we are concerned with knowing the mean of the population w

Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study

The population mean in this case is simply the mean of the swimming times of the 460 swimmers

There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study

So conclusively, the population mean w is simply the mean of the total 460 swimmers

Answer:

Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second

Explanation:

Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg

Velocity of the car is 2.5 meter per second

Since formula to calculate the momentum of an object is,

p = mv

Where, p = momentum of the object

m = mass of the object

v = velocity of the object

By substituting these values in the formula,

p = [tex](6.3\times 10^{-3})\times 2.5[/tex]

  = [tex]1.575\times 10^{-2}[/tex] Kg meter per second

Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.

Answer:

A, C

Explanation:

Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest

A. Replacing the string with a thicker string.

Thicker strings have more density. The more density the string has, the lower the sound.

Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:

For:

[tex]f=\frac{v}{\lambda}[/tex]

[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]

See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.

C. Doubling the length of the string.

Because the length of the string is inversely proportional to the frequency.

The longer the string, the lower the frequency.

So, if we double string, we'll hear lower sounds in any string instrument

--

In short,  for A, and C  We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W

Answer:

a) 43.98 V

b) E = 21.37 MJ

Explanation:

Parameters given:

Length of cable = 180 km = 180000 m

Diameter of cable = 11 cm = 0.11 m

Radius = 0.11 / 2 = 0.055 m

Current, I = 135 A

a) To find the potential drop, we have to find the voltage across the wire:

V = IR

=> V = IρL / A

where R = resistance

L = length of cable

A = cross-sectional area

ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm

Therefore:

V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)

V = 43.98 V

The potential drop across the cable is 43.98 V

b) Electrical energy is given as:

E = IVt

where t = time taken = 1 hour = 3600 s

Therefore, the energy dissipated per hour is:

E = 135 * 43.98 * 3600

E = 21.37 MJ (mega joules, 10^6)

Answer:

t = 3.4 s

The box will come to rest in 3.4 s

Explanation:

For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma

Frictional Force = μR = μW = μmg

Therefore,

ma = μmg

a = μg

where,

a = acceleration of box = ?

μ = coefficient of sliding friction = 0.45

g = 9.8 m/s²

Therefore,

a = (0.45)(9.8 m/s²)

a = -4.41 m/s²  (negative sign due to deceleration)

Now, for the time to stop, we use first equation of motion:

Vf = Vi + at

where,

Vf = Final Speed = 0 m/s (since box stops at last)

Vi = Initial Speed = 15 m/s

t = time to stop = ?

Therefore,

0 m/s = 15 m/s + (-4.41 m/s²)t

(-15 m/s)/(-4.41 m/s²) = t

t = 3.4 s

The box will come to rest in 3.4 s

Answer:

clockwise

Explanation:

when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.

Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same  polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.

The direction of the induced current in the bottom ring is in the clockwise direction.

The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.

Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.

Learn more about the concept of induced current here:

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Answer:

Explanation:

Kinetic energy at the height = 1/2 m v²

= 1/2 x 750 x 20²

= 150000 J

Its potential energy = mgh

= 750 x 9.8 x 5

=36750 J

Total energy = 186750 J

Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy

If v be its velocity at the bottom

1/2 m v² = 186750

v = √498

= 22.31 m /s

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

[tex]D = 0.5 c \rho A v^2[/tex]

Object 1 has twice the diameter of object 2.

If [tex]d_2 = d[/tex]

[tex]d_1 = 2d[/tex]

Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]

Area of object 1:

[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]

Since all other parameters are still the same except the drag coefficient:

For object 1:

[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]

For object 2:

[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]

Since the drag force for the two objects are the same:

[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

Answer:

the internal energy of the mixture at final state = 238kJ/kg

Explanation:

Given

V= 0.6m³

m=5kg

R=0.287kJ/kg.K

T=320 K

from ideal gas equation

PV = nRT

where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.

Recall, mole = mass/molar mass

attached is calculation of the question.

Answer:

The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"

Explanation:

length of the copper wire:

L= 62.9 cm

r is the radius of the loop then:

[tex]r=\frac{L}{2 \pi}\\[/tex]

  [tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]

area of the loop Is:

[tex]A_L= \pi r^2[/tex]

     [tex]=100.2001\times 3.14\\\\=314.628[/tex]

change in magnetic field is:

[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]

then the induced emf is:  [tex]e = A_L \times \frac{dB}{dt}[/tex]

                                              [tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]

resistivity of the copper wire is: [tex]\rho =[/tex]  1.69 × 10-8Ω·m

diameter d = 1.15mm

radius (r) = 0.5mm

               [tex]= 0.5 \times 10^{-3} \ m[/tex]

hence the resistance of the wire is:

[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]

   [tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]

Power:

[tex]P=\frac{e^2}{R}[/tex]

[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]

The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

Answer:

t = 0.31s

Explanation:

In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:

[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]

A: amplitude

v_max = 1.38m/s

a_max = 6.83m/s^2

w: angular frequency

From the previous equations you can obtain the angular frequency w.

You divide vmax and amax, and solve for w:

[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]

Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.

You calculate the period  by using the information about the angular frequency:

[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]

Then the required time is:

[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]

Answer: lower

There are a number of factors that can affect the coefficient of friction, including surface conditions.

Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively

Answer:

Increasing population

Explanation:

As we can see that the death rate is decreasing while at the same time the birth rate is increasing due to which it increased the population that directly impact the planet so great

Day by day the population of the villages, cities, states, the country is increasing which would create a direct human impact on the planet  

Therefore the increasing population is the one and single most important reason

Answer:

  a = 1.15 10³ g

Explanation:

For this exercise we will use the relations of the centripetal acceleration

     a = v² / r

where is the linear speed of the rotor and r is the radius of the rotor

let's use the relationships between the angular and linear variables

          v = w r

let's replace

          a = w² r

let's reduce the angular velocity to the SI system

        w = 550 rev / min (2pi rad / 1 rev) (1 min / 60 s)

         w = 57.6 rad / s

let's calculate

       a = 57.6²  3.4

       a = 1.13 10⁴ m / s²

To calculate this value in relation to g, let's find the related

       a / g = 1.13 10⁴ / 9.8

       a = 1.15 10³ g

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  [tex]U =\frac{Qmax^2}{C}[/tex]

So conclude C doesn't change substantially as well as,

When,

⇒  [tex]Q=\frac{Qmax}{2}[/tex]

⇒  [tex]Q^2=\frac{Qmax^2}{4}[/tex]

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.