PHYSICS COLLEGE

How long is a bus route across a small town

Answers

Answer 1

Answer:

4 kilometers

Explanation:

Answer 2

Answer:

Final answer:

The length of a bus route in a small town can greatly depend on the specifics of the town and the route. It can range from a couple miles in a very small town to 20 miles or more for larger towns.

Explanation:

The length of a bus route across a small town can vary greatly depending on the size of the town and the specifics of the bus route. In a very small town, the bus route might only be a mile or two long. For larger towns, it could easily be 10-20 miles, or more. If you know the specifics of the route (streets it travels along, the number of stops, etc.), you could use a tool like Go_gle Maps to calculate an approximate distance.

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If a microwave oven produces electromagnetic waves with a frequency of 2.30 ghz, what is their wavelength?

Answers

Answer: wavelength is $<strong> 1.30 * 10^8 nm.</strong>$

The frequency of the microwave is, f = 2.30 GHz.

To Find frequency use the formula:

$c=f$ λ

Where, c is the speed of electromagnetic wave or light. f is the frequency, and λ is the wavelength of light.

Rearranging, $\lambda = (c)/(f)$

Plug in the values,

$\lambdam = (3 * 10^8 m/s)/(2.30 GHz(10^9 Hz)/(1 GHz))=0.130 m(10^9 nm)/(1 m) = 1.30 * 10^8 nm.$

$\lambda = (c)/(\nu) = (3 \cdot 10^8)/(2.3 \cdot 10^9) = (3)/(23) \approx 0.130435 \approx 0.13 \ m.$

A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

Answers

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

$\int\ {E} \, dA = E*A = E*\pi *R^(2)$

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?

Answers

Answer:

The value of the magnetic field is 2.01 T when Hall voltage is 1.735 mV

Explanation:

The frequency of the cyclotron can help us find the magnitude of the magnetic field, thus then we can compare the effect of increasing Hall voltage on the probe.

Magnetic field magnitude at initial Hall voltage.

The cyclotron frequency can be written in terms of the magnetic field magnitude as follows

$f = \cfrac{qB}{2\pi m}$

Solving for the magnetic field.

$B = \cfrac{2\pi mf}q$

Thus we can replace the given information but in Standard units, also remembering that the mass of a proton is $m_p=1.67 * 10^(-27) kg$ and its charge is $q_p=1.6 * 10^(-19) C$ .

So we get

$B = \cfrac{2\pi * 1.67 * 10^(-27) kg * 9.7 * 10^6 Hz}{1.6 * 10^(-19)C}$

$B =0.636 T$

We have found the initial magnetic field magnitude of 0.636 T

Magnetic field magnitude at increased Hall voltage.

The relation given by Hall voltage with the magnetic field is:

$V_H =\cfrac{R_HI}t B$

Thus if we keep the same current we can write for both cases:

$V_(H1) =\cfrac{R_HI}t B_1\nV_(H2) =\cfrac{R_HI}t B_2$

Thus we can divide the equations by each other to get

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{\cfrac{R_HI}t B_1}{\cfrac{R_HI}t B_2}$

Simplifying

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{ B_1}{ B_2}$

And we can solve for $B_2$

$B_2 =B_1 \cfrac{V_(H2)}{V_(H1)}$

Replacing the given information we get

$B_2= 0.636 T * \left(\cfrac{1.735 mV}{0.549 mV} \right)$

We get

$\boxed{B=2.01\, T}$

Thus when the Hall voltage is 1.735 mV the magnetic field magnitude is 2.01 T

The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

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

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

Answers

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

$\Rightarrow x=ut+0.5at^2 \n\Rightarrow 11=4.3t+0.5(4.6)t^2\n\Rightarrow 2.3t^2+4.3t-11=0\n\Rightarrow (t-1.4435)(t+3.3131)=0\n\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\n$