PHYSICS COLLEGE

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a wavelength of 3.5 cm?Express your answer in GHz and using two significant figures.
f = GHz
Microwave signals are beamed between two mountaintops 52 km apart. How long does it take a signal to travel from one mountaintop to the other?
Express your answer in ms and using two significant figures.
t = ms

Answers

Answer 1

Answer:

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =(c)/(\lambda) =(3e8m/s)/(0.035m) = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =(d)/(t) (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = (d)/(c) = (52e3m)/(3e8m/s) = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

⇒ t = 0.2 ms (with two significative figures)

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A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 349 K, what is its radius at liftoff

Answers

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $(4)/(3) \pi r^(3)$

volume of balloon = $(4)/(3)$ x 3.142 x $24^(3)$ = 57913.34 m^3

using the gas equation,

$(P1V1)/(T1)$ = $(P2V2)/(T2)$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$(0.03*57913.34)/(200)$ = $(1*V2)/(349)$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $(4)/(3) \pi r^(3)$ = $(4)/(3)$ x 3.142 x $r^(3)$ = 3031.76

4.19 $r^(3)$ = 3031.76

$r^(3)$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb

Answers

To solve this problem we need to use the emf equation, that is,

$E=m(dI)/(dT)$

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

$M=(E)/((dI)/(dT))\nM=(1.6*10^(-3))/(0.245)=6.53*10^(-3)H$

Solving for b) we need the FLux through each turn, that is

$\Phi=(MI)/(N)$

Where N is the number of turns in the second coil

$\Phi=(6.53*10^(-3)*1.25)/(22)=3.71*10^(-4)Wb$

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

Answers

Answer:

The change in momentum is $\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A = <5600, 7200,0>$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = <-4000 , -780 , 0> N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)$

$\Delta p = <-4000 , -780 , 0> * (0.3)$

$\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?

Answers

Answer:

velocity of the river is equal to 0.56 m/s

Explanation:

given,

velocity of swimmer w.r.t still water = 1 m/s

width of river = 73 m

he arrives to the point = 41 m

$times = (distance)/(speed)$

$times = (73)/(1)$

t = 73 s

$velocity = (distance)/(time)$

= $(41)/(73)$

= 0.56 m/s

velocity of the river is equal to 0.56 m/s

What is the answer to life and everything in the universe?

Answers

The answer to life and the universe is 42

Determine the tension in the string that connects M2 and M3.

Answers

thereforemass m1=4.8kg and the tension

in the horizontalspring T2=10N.

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To determine the tension in the string that connects M2 and M3, we can follow these steps:

Step 1: Identify the necessary variables. Given data (for example) could be:
- Mass of M2, which is 5 kg
- Mass of M3, which is 10 kg
- The acceleration due to gravity, which is approximately 9.8 m/s²
- The angle at which the string pulls on M2, which is 30 degrees
- Assume the system is in equilibrium, meaning there is no net acceleration, so the acceleration is 0 m/s²

Step 2: Calculate the weight of M3, which is its mass times the acceleration due to gravity. This is because weight is the force exerted by gravity on an object, which equals the object's mass times the acceleration due to gravity.

For M3, this calculation would be M3 * g = 10 kg * 9.8 m/s² = 98 N (Newtons).

Step 3: Determine the force exerted by M2 that acts along the line of the string. This won't be the full weight of M2, because the string pulls at an angle. This component of the force can be calculated using the sine of the angle, because sine gives us the ratio of the side opposite the angle (here, the force along the string) to the hypotenuse (here, the full weight of M2) in a right triangle.

The horizontal component of the force of M2 is then M2 * g * sin(30deg) = 5 kg * 9.8 m/s² * sin(30deg) = 24.5 N.

Step 4: The tension in the string is the force M3 exerts on it, which is its weight, minus the component of M2's weight that acts along the string. This is because M2 and M3 are pulling in opposite directions, so they subtract from each other.

The tension in the string is then the weight of M3, 98 N, minus the horizontal (along the string) component of M2's weight, 24.5 N.

So, the tension in the string is 98 N - 24.5 N = 73.5 N.

This is the force that the string needs to exert in order to keep M2 and M3 connected and in equilibrium.

Learn more about Tension in a string here:

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