What do radio waves and microwaves have in common?

Physics

2 answers:

barxatty [35]3 years ago

7 0

Answer:

Both are found next to visible light on the electromagnetic spectrum.

i know because i just took the test trust me i choose Both are found next to visible light on the electromagnetic spectrum. and got it wrong it said the answer was Both are found next to visible light on the electromagnetic spectrum.

Explanation:

forsale [732]3 years ago

3 0

Answer:

Both are at the side of the spectrum that has the lowest frequency.

Explanation:

We know that the electromagnetic spectrum is the set of several waves that are arranged in order of increasing wavelength of decreasing frequency. It consists of many waves like,

gamma rays, x- rays, ultraviolet rays, visible, infrared rays, microwave, radio waves.

Radio wave has the longest wavelength while gamma rays has maximum wavelength. From the attached figure, it is clear that the frequency of radio wave is least and that of microwave is second least frequency.

Hence, the correct option is (c).

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a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitti

poizon [28]

Ans: R = Ball Travelled = 92.15 meters.

Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.

Along x-axis, equation would be:
$x = x_o + v_o_xt + \frac{at^2}{2}$

Since there is no acceleration along x-direction; therefore,
$x = x_o + v_o_xt$

Since $v_o_x = v_ocos \alpha$ and $x_o$ =0; therefore above equation becomes,

$x = v_ocos \alpha t$ --- (A)

Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,

=> $y = y_o + v_o_yt - \frac{gt^2}{2}$
=> $t = \frac{2v_o_y}{g}$

Since $v_o_y = sin \alpha$ ; therefore above equation becomes,
$t = \frac{2v_osin \alpha }{g}$

Put the value of t in equation (A):

(A) => $x = v_ocos \alpha \frac{2v_osin \alpha }{g}$

Where x = Range = R, and $2sin \alpha cos \alpha = sin(2 \alpha )$ ; therefore above equation becomes:

=> $R = (v_o)^2 *\frac{sin(2 \alpha )}{g}$

Now, as:
$v_o = 31 m/s$

and $\alpha = 35$ °
and g = 9.8 m/(s^2)

Hence,
$R = (31)^2 *\frac{sin(2 *35 )}{9.8}$

Ans: R = 92.15 meters.

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