PHYSICS COLLEGE

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.90 keV and a current of 4.95 mA produced by the generator. (a) What is the speed of the protons (in m/s)?

Answers

Answer 1

Answer:

Answer:

603383.67253 m/s

Explanation:

m = Mass of proton = $1.67* 10^(-27)\ kg$

K = Kinetic energy = 1.9 keV

$1\ ev=1.6* 10^(-19)\ J$

Velocity of proton is given by

$v=\sqrt{(2K)/(m)}\n\Rightarrow v=\sqrt{(2* 1.9* 10^3* 1.6* 10^(-19))/(1.67* 10^(-27))}\n\Rightarrow v=603383.67253\ m/s$

The speed of the protons is 603383.67253 m/s

Related Questions

The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)
Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.
Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.
An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?
An underwater sound source emits waves of frequency 30 kHz in all directions. How does the intensity of the waves (in Watts/m2) vary with distance r from the source?a) 1/r^3b) 1/r^2c) 1/rd) None of above

A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A package is launched upward, from a point on a roof 10 m above the ground. The initial velocity of the package is 50.5 m/s. Consider all quantities as positive in the upward direction. Does Jim Bond have a chance to catch the package? (calculate how high will it go)

Answers

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V. Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Answers

Answer:

U_eq = 1.99 * 10^(-10) J

Explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12 C^2 / N .m

Equations used:

U = 0.5 C*V^2 .... Eq 1

C = e_o * k*A /d .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d

U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J

A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat

Answers

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

$v=\lambda f=\lambda(1)/(T)=(110m)(1)/(8.3s)=13.25(m)/(s)$

the relative velocity is:

$v'=13.25m/s-5m/s=8.25(m)/(s)$

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

$x'=v't=(8.25m/s)(20s)=165m$

Next, you use the general for of a wave:

$f(x,t)=Acos(kx-\omega t)=Acos((2\pi)/(\lambda)x-\omega t)$

you take the amplitude as 2.0/2 = 1.0m.

$\omega=(2\pi)/(T)=(2\pi)/(8.3s)=0.75(rad)/(s)$

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

$f(165,20)=1.0m\ cos((2\pi)/(110m)(165)-(0.75(rad)/(s))(20s))\n\nf(165,20)=0.99m$

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it takethe ball to reach the highest point?

Answers

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

Which is true about the radiation force of light shining on a surface? The force is greater if the light reflects back along its incident path than in some other direction. The force is greater if the light is absorbed instead of being reflected. The force is greater if the light is reflected in some direction other than back along the incident path.

Answers

Answer:

The force is greater if the light is absorbed instead of being reflected

Explanation:

Light could either be reflected or absorbed. Reflection takes place where light is concentrated back to another surface whereas absorption takes place when light is incorporated into a surface thereby providing kinetic energy. The kinetic energy produced by absorption provides more force than reflection which just involves concentrating back to another surface.

A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.5 m/sm/s , a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic

Answers

Answer:

$v = 8.333\,(m)/(s)$

Explanation:

The system mosquito-raindrop is described by the Principle of Momentum Conservation:

$(50\cdot m)\cdot (8.5\,(m)/(s))+m\cdot (0\,(m)/(s) ) = 51\cdot m \cdot v$

$425 = 51\cdot v$

The final speed is:

$v = 8.333\,(m)/(s)$

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