A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y-direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line. The electric field should be in the ______ direction.

Answers

Answer 1

Answer:

Answer:

Negative z-direction.

Explanation:

We need to determine the direction of the magnetic force. Since the velocity of the proton is in the positive x direction, and the magnetic field is in the positive y direction, we know by the vectorial formula $F=q(v* B)$ (or, alternatively, with the left hand rule) that the magnetic force points in the positive z-direction (also taking into account that the charge is positive), so the electric field should be in the negative z-direction to balance it.

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. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.You should obtain e/m = 2V/(B^2)(r^2)3. The magnetic field on the axis of a circular current loop a distance z away is given byB = mu I R^2 / 2(R^2 + z^2)^ (3/2)where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/Rwhere B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters
Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3 , is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, i

Answers

Final answer:

To find the final pressure, use the ideal gas law equation PV = nRT, where P is the initial pressure, V is the initial volume, n is the number of moles of gas, R is the gas constant, and T is the initial temperature. Rearrange the equation and plug in the given values to find that the final pressure is 3.33 bar.

Explanation:

To find the final pressure, we can use the ideal gas law equation: PV = nRT, where P is the initial pressure, V is the initial volume, n is the number of moles of gas, R is the gas constant, and T is the initial temperature.

Since the volume and the amount of air are constant, we can rearrange the equation to solve for the final pressure:

P2 = P1 * (T2 / T1),

where P2 is the final pressure, T2 is the final temperature, and T1 is the initial temperature.

By plugging in the values from the problem, we can find that the final pressure is 3.33 bar.

Learn more about Ideal Gas Law here:

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Light has wavelength 600 nm in a vacuum. it passes into glass, which has an index of refraction of 1.5. what is the frequency of the light inside the glass

Answers

Light has wavelength 600 nm in a vacuum ,the frequency of the light is 2 × $10^(-13)$ Hz.

What is wavelength?

The separation between such a wave motion's crests and troughs would be known as the wavelength of photons.

What is frequency ?

The total number of waves that pass a specific location in a predetermined amount of time is known as frequency.

Calculation of frequency

Given data:

wavelength = 600 nm = 600 × $10^(-9)$ m

index of refraction = 1.5.

Frequency can be calculated by using the formula:

v = f × wavelength

f = wavelength / v

Where, f = Frequency , v is velocity.

put the given data in above equation.

f = wavelength / v

f = 600 × $10^(-9)$ m / 3 × $10^(8)$

f = 200 × $10^(-15)$ .

f = 2 × $10^(-13)$

Therefore, the frequency of the light is 2 × $10^(-13)$ Hz.

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v = f lambda

in vac ... 3X10^8 = 600x10^-9xf

in glass speed slower, poss 2/3 that of vacuum

A proton and an alpha particle (helium nucleus consisting of two protons and two neutrons) are accelerated from rest across the same potential difference. Assume the proton mass and the neutron mass are roughly the same and neglect any relativistic effect. Compared to the final speed of the proton, the final speed of the alpha particle is?1. less by a factor of 22. less by a factor of √ 23. less by a factor of 44. greater by a factor of 25. the same

Answers

Answer:

option B

Explanation:

we know,

change in energy is equal to

$W = (1)/(2)m(v^2 - u^2)$

$W = (1)/(2)m(v^2 - 0^2)$

$W = (1)/(2)m v^2$

$q = (1)/(2)m v^2$

proton mass and the neutron mass are roughly the same

so,

$q \alpha m v^2$

now,

$(q_p)/(q_(\alpha)) = (m_p v_p^2)/(m_(\alpha)v_(\alpha)^2)$

$(q_p)/(q_(\alpha)) = (m_p v_p^2)/(2 m_pv_(\alpha)^2)$

we know,

mass of alpha particle is four times mass of the mass of proton.

mα = 4 m_p

$(e)/(2e) = ( v_p^2)/(4 v_(\alpha)^2)$

$( v_p^2)/(v_(\alpha)^2) = 2$

$v_(\alpha)^2 =( v_p^2)/(2)$

$v_(\alpha)=( v_p)/(√(2))$

less by a factor of √2

Hence, the correct answer is option B

As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None

Answers

Answer:

B- Velocity

Explanation:

This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_(0)(1-e^{(-t)/(RC) } )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) }) \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )$

Taking natural log on both sides,

$e^{(-t)/(RC) } = 1- (V)/(V_(0) ) \n(-t)/(RC) = ln(1-(V)/(V_(0) ) )\nt = -RCln(1 - (V)/(V_(0) ))$

$t = -RC ln (1-(V)/(V_(0) ))$ (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

= $((8.85*10^(-12))) (20*10^(-4)) )/(1*10^(-3) )$

= 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

$t = -RC ln (1-(V)/(V_(0) ))$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).

Answers

Answer:

k = 15.62 MN/m

Explanation:

Given:-

- The viscous damping constant, c = 1.8 KNs/m

- The floor oscillation magnitude, Yo = 3 mm

- The frequency of floor oscillation, f = 18 Hz.

- The combined weight of the grinding machine and the wheel, W = 4200 N

- Two springs of identical stiffness k are attached in parallel arrangement.

Constraints:-

- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )

Find:-

What is the minimum required stiffness of each of the two springs as per the constraints given.

Solution:-

- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

$y ( t ) = Y_o*sin ( w*t )$

Where,

Yo : The amplitude of excitation = 3 mm

w : The excited frequency = 2*π*f = 2*π*18 = 36π

- The harmonic excitation of the floor takes the form:

$y ( t ) = 3*sin ( 36\pi *t )$

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

$m*(d^2x)/(dt^2) + c*(dx)/(dt) + k_e_q*x = k_e_q*y(t) + c*(dy)/(dt)\n\n(m)/(k_e_q)*(d^2x)/(dt^2) + (c)/(k_e_q)*(dx)/(dt) + x = y(t) + (c)/(k_e_q)*(dy)/(dt)$

Where,

m: The combined mass of the rigid body ( wheel + grinding wheel body) c : The viscous damping coefficient

k_eq: The equivalent spring stiffness of the system ( parallel )

x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

$w_n = \sqrt{(k_e_q)/(m) } , \n\np = (c)/(2√(k_e_q*m) ) = (1800)/(2√(k_e_q*428.135) ) = (43.49628)/(√(k_e_q) )$

Where,

w_n: The natural frequency

p = ζ = damping ratio = c / cc , damping constant/critical constant

- The Equation of motion becomes:

$(1)/(w^2_n)*(d^2x)/(dt^2) + (2*p)/(w_n)*(dx)/(dt) + x = y(t) + (2*p)/(w_n)*(dy)/(dt)$

- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

$X_s_s = X_o*sin ( wt + \alpha )$

Where,

X_o : The amplitude of the steady-state vibration.

α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

$X_o = Y_o*\sqrt{(1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2) }$

Where,

r = Frequency ratio = $(w)/(w_n) = \frac{36*\pi }{\sqrt{(k_e_q*g)/(W) } } = \frac{36*\pi }{\sqrt{(k_e_q)/(428.135) } } = (36*\pi*√(428.135) )/(√(k_e_q) )$

- We will use the one of the constraints given to limit the amplitude of steady state oscillation ( Xo ≤ 10 mm ):

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):

$((X_o )/(Y_o))^2 \geq (1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2)\n\n((X_o )/(Y_o))^2 \geq (1+ ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)/(( 1 - ((36*\pi*√(428.135) )/(√(k_e_q) ))^2)^2 + ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)\n\n$

$((X_o )/(Y_o))^2 \geq ( 1 + (41442858448.85813)/(k_e_q^2 ))/([ 1 - ((5476277.91201 )/(k_e_q) )]^2 + (41442858448.85813)/(k_e_q^2 ) )}\n\n((X_o )/(Y_o))^2 \geq ( (k_e_q^2 + 41442858448.85813)/(k^2_e_q ))/([ ((k_e_q - 5476277.91201)^2 )/(k_e_q^2) ] + (41442858448.85813)/(k_e_q^2 ) )}\n$

$((X_o )/(Y_o))^2 \geq ( k_e_q^2 + 41442858448.85813)/( (k_e_q - 5476277.91201)^2 +41442858448.85813 )}\n\n((10 )/(3))^2 \geq ( k_e_q^2 + 41442858448.85813)/( k^2_e_q -10952555.82402*k_e_q +3.00311*10^1^3 )}\n\n\n10.11111*k^2_e_q -121695064.71133*k_e_q +3.33637*10^1^4 \geq 0$

- Solve the inequality ( quadratic ):

$k1_e_q \geq 7811740.790197058 (N)/(m) \n\nk2_e_q \leq 4224034.972855095 (N)/(m)$

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

$k_e_q = (k^2)/(2k) = (k)/(2)$

- Therefore,

$k1 \geq 7811740.790197058*2 = 15.62 (MN)/(m) \n\nk2 \leq 4224034.972855095*2 = 8.448 (MN)/(m)$

- The minimum stiffness of spring is minimum of the two values:

k = 15.62 MN/m

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