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AnnyKZ [126]
3 years ago
6

A police radar gun uses X-band microwave radiation at a frequency of 12.2 GHz. Microwaves travel at the speed of light, or 3x108

m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car. Prof Chesnutt is driving down the road, and her speed is measured by a friendly officer of the law sitting by the side of the road.
what is the resulting beat frequency that the radar gun detects?
Physics
1 answer:
quester [9]3 years ago
6 0

Answer:

A police radar gun uses X-band microwave radiation at a frequency of 13.1 GHz. Microwaves travel at the speed of light, or 3x108 m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car.

a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

ANSWER: 2533 Hz

Explanation:

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1600 AM broadcasts radio waves with wavelengths<br> of about 187.37 m. Convert this to miles.
nikklg [1K]

Answer:

The wavelength in miles is <u>0.1165 miles</u>.

Explanation:

Given:

Wavelength of the radio wave is 187.37 m.

Now, the wavelength is given in meters.

We need to convert the wavelength from meters to miles.

In order to convert meters to miles, we have to use their conversion factor.

We know that,

1 meter = \frac{1}{1609}\ miles

Therefore, the conversion factor is given as:

CF=\frac{1}{1609}\ miles\ per\ meter

So, the wavelength in miles is given as:

Wavelength=\textrm{Wavelength in meters}\times CF\\\\Wavelength=187.37\ m\times \frac{\frac{1}{1609}\ miles}{1\ m}\\\\Wavelength=\frac{187.37}{1609}\ miles\\\\Wavelength=0.1165\ miles

Hence, the wavelength in miles is 0.1165 miles.

7 0
3 years ago
A tow truck exerts a net force of 1000 N on a 600 kilogram car What’s the cars acceleration
Mice21 [21]

Answer:

a = 1000/600= 5/3 m/s^2

5 0
3 years ago
6.
yaroslaw [1]

Answer:

12 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 7.6 kg

Distance (d) = 6 m

Velocity (v) = 5 m/s

Force (F) = 2 N

Workdone (Wd) =.?

Workdone can be defined as the product of force and distance moved in the direction of the force. Mathematically, it is expressed as:

Workdone = Force × distance

Wd = F × d

With the above formula, we can obtain the workdone as follow:

Distance (d) = 6 m

Force (F) = 2 N

Workdone (Wd) =.?

Wd = F × d

Wd = 2 × 6

Wd = 12 J

Thus, the workdone is 12 J

6 0
3 years ago
Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm
Ghella [55]

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

3 0
3 years ago
Suppose a straight wire with a length of 2.0 m runs perpendicular to a magnetic field with a magnitude of 38 T. What current wou
iragen [17]

Answer:

9.67 A

Explanation:

The weight of a student with a mass of m = 75 kg is:

W=mg=(75 kg)(9.8 m/s^2)=735 N

where g=9.8 m/s^2 is the acceleration due to gravity.

We want the magnetic force on the wire to be equal to this weight. The magnetic force on the wire is

F=ILB sin \theta

where

I is the current in the wire

L = 2.0 m is the length of the wire

B = 38 T is the magnetic field

\theta=90^{\circ} is the angle between the direction of B and L

Since we want W=F, we can write

ILB sin \theta=W

And we can solve it to find the current I:

I=\frac{W}{BLsin\theta}=\frac{735 N}{(38 T)(2.0 m)(sin 90^{\circ})}=9.67 A

8 0
3 years ago
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