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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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Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward
Radda [10]

Answer:

D = 25 miles

Explanation:

To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.

Since both bicyclists collide, we know that Xa=Xb, so:

Xa = V*t = 10*t     and    Xb = 20 - V*t = 20 - 10*t

10*t = 20 - 10*t      Solving for t:

t = 1 hour  Now we can calculate the distance for the bee:

D = Vbee * t = 25 * 1 = 25 miles

6 0
4 years ago
In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 609 nm.
Ray Of Light [21]

Answer:

\Delta x = 3.65 \mu m

Explanation:

As we know that the sixth order maximum will have path difference given as

\Delta x = N\lambda

here we know that

N = order of maximum

\lambda = 609 nm

now we have

N = 6

so we know that

\Delta x = 6(609 nm)

\Delta x = 3.65 \mu m

6 0
3 years ago
Question 26 suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of
ololo11 [35]

<span>The answer is 6 kg the mass of the second object. By using Inversely proportional formula it means that (14 kg) (3 m/s</span>²<span>) = M (7 m/s</span>²<span>). Where M is the mass of the second object. For the Newton’s second law of motion formula which is: Force = mass x acceleration, we have:</span>

<span>F = (14 kg) (3 m/s</span>²<span>) = 42 N</span>

 

Therefore:

<span>42 N = M (7 m/s</span>²)

<span>M = (42 N) / (7 m/s</span>²<span>)</span>

M = 6 kg mass of the second object

4 0
3 years ago
In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move
lara31 [8.8K]

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.  

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

8 0
3 years ago
9. A ball is thrown straight up with an initial speed of 30 m/s. How long will it take to reach the top of its trajectory, and h
ziro4ka [17]

Answer:

1.) Time t = 3.1 seconds

2.) Height h = 46 metres

Explanation:

given that the initial velocity U = 30 m/s

At the top of the trajectory, the final velocity V = 0

Using first equation of motion

V = U - gt

g is negative 9.81m/^2 as the object is going against the gravity.

Substitute all the parameters into the formula

0 = 30 - 9.81t

9.81t = 30

Make t the subject of formula

t = 30/9.81

t = 3.058 seconds

t = 3.1 seconds approximately

Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.

2.) The height it will go can be calculated by using second equation of motion

h = ut - 1/2gt^2

Substitutes U, g and t into the formula

h = 30(3.1) - 1/2 × 9.8 × 3.1^2

h = 93 - 47.089

h = 45.911 m

It will go 46 metres approximately high.

6 0
3 years ago
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