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3 years ago

What physics principle is used in radar guns to find the speeds of tennis balls and baseballs at sporting events?

Physics

1 answer:

Luda [366]3 years ago

6 0

<h2>Answer: Doppler effect </h2>

Explanation:

A radar gun (also known as a Doppler radar) uses the Doppler effect when measuring "return echoes" after having sent a microwave signal (a type of electromagnetic radiation).

In this context the Doppler effect consists of the change in a wave perceived frequency when the emitter of the waves, and the observer move relative to each other.

In the case of radars, a microwave signal is sent to a target (the tennis or baseball in this case) and then is reflected after "hitting" the target, so that the radar system measures this difference between the sent signal and the reflected signal.

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A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

Explanation:

For an open pipe, the velocity is given by

$v=\frac {2Lf}{n}$

Making L the subject then

$L=\frac {nV}{2f}$

Where f is the frequency, L is the length, n is harmonic number, v is velocity

Substituting 1 for n, 320 Hz for f and 331 m/s for v then

$L=\frac {1*331}{2*320}=0.5171875\approx 0.52m$

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, $f2=2\times \frac {v}{2L}$ and $f3=3\times \frac {v}{2L}$

$f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz$

(c)

When v=367 m/s then

$f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz$

5 0

3 years ago

When a force of 450N pushes on a 20kg box as

ehidna [41]

Answer:

ma+mgsinh0+f=F∴(25)(0.75)+(25)(10)sinh0+μkN=F∴18.75+(250)(0.6h)+μk(mgcosh0=F⟹18.75+150+μk((25)(10)(0.76))=500∴168.75+μk(190)=500⟹μk(190)=331.25⟹μk=1.74

Explanation:

7 0

2 years ago

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the

allsm [11]

Answer:

r = 6.5*10^-3 m

Explanation:

I'm assuming you meant to ask the diameters of the disk, if so, here's it

Given

Quantity of charge on electron, Q = 1.4*10^9

Electric field strength, e = 1.9*10^5