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3 years ago

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).

You might be interested in

Calculate the current through a 3.0 Ď‰ resistor with a voltage of 9.0 v across it.

gulaghasi [49]

Answer: 3 A

Explanation:

According to Ohm's law:

$V=R.I$

Where:

$V=9 V$ is the voltage

$R=3\Omega$ is the resistance of the resistor

$I$ is the electric current (the value we want to find)

Isolating $I$ :

$I=\frac{V}{R}$

$I=\frac{9 V}{3\Omega}$

Finally:

$I=3 A$

7 0

2 years ago

A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap

vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

$W = mgH$

here we know that

$mg = 5350 N$

also we know that

H = 75 m

now we have

$W = 5350 \times 75$

$W = 401250 J$

7 0

3 years ago

What formula is used to find an objects acceleration

ololo11 [35]

Acceleration = velocity / time.

7 0

3 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago

A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats

Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$

Where:

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

$\lambda = \frac{c}{f}$

Where:

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

$\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m$

Hence, the distance is:

$d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m$

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

I hope it helps you!

7 0

2 years ago