An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of

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Answer 1

Answer:

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = (1)/(2)mv^2$

Making v the subject

$v = \sqrt{[(2 \Delta V * q )/(m)] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{(2 * 5.9 *10^3 * 1.60218*10^(-19) )/(9.10939 *10^(-31]) )$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $(mv^2)/(r)$ ] and this is mathematically represented as

$Bqv = (mv^2)/(r)$

making B the subject

$B = (mv)/(rq)$

r is the radius with a value = 5.4cm = $= (5.4)/(100) = 5.4*10^(-2) m$

Substituting values

$B = (9.1039 *10^(-31) * 4.556 *10^7)/(5.4*10^-2 * 1.60218*10^(-19))$

$= 0.0048 T$

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A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answers

To solve this problem it is necessary to apply the concepts related to Newton's second law and the equations of motion description for acceleration.

From the perspective of acceleration we have to describe it as

$a = (\Delta v)/(\Delta t)$

Where,

$\Delta v$ = Velocity

$\Delta t$ = time

At the same time by the Newton's second law we have that

F = ma

Where,

m = mass

a = Acceleration

Replacing the value of acceleration we have

$F = m ((\Delta v)/(\Delta t))$

Our values are given as,

$m = 55*10^(-3)Kg$

$v = 46m/s$

$t = 0.00340s$

Replacing we have,

$F = m ((\Delta v)/(\Delta t))$

$F = (55*10^(-3))((46)/(0.00340))$

$F = 744.11N$

Therefore the magnitude of the average force exerted on the ball by the club is 744.11N

A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answers

Answer:

E = 0.13 J

Explanation:

At resonance condition we have

$\omega = \sqrt{(1)/(LC)}$

$\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}$

$\omega = 1217.2 rad/s$

now if the frequency is double that of resonance condition then we have

$x_L = 2\omega L$

$x_L = 2(1217.2)(13.5* 10^(-3))$

$x_L = 32.86 ohm$

now we have

$x_c = (1)/(2(1217)(50* 10^(-6)))$

$x_c = 8.22 ohm$

now average power is given as

$P = i_(rms)V_(rms)* (R)/(z)$

$P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))$

$P_(avg) = 50.67 Watt$

Now time period is given as

$T = (2\pi)/(\omega)$

so total energy consumed is given as

$E_(avg) = 50.67((2\pi)/(2(1217.2)))$

$E = 0.13 J$

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Answers

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

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For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(700-550)/(550)*3*10^(8)\n v=8.1818*10^(7)m/s$

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Answers

Answer:

the correct option is C

Explanation:

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

The following three appliances are connected to a 120 volt house circuit: (i) a toaster, 1200 Watts, (ii) a coffee pot, 750 Watts, and (iii) a microwave, 600 Watt. If all operate at the same time, what total current would they draw

Answers

Answer:

They would draw a total of 21.25 amperes

Explanation:

The total power consumed is

1200 W+ 750 W + 600 W= 2550 Watts

The formula relating the power consumed, the voltage and the current is given as

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given that the voltage supply is 120V

$2550=I*120\n\I=(2550)/(120) \n\nI= 21.25amps$

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