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3 years ago

A similarity between radio waves and microwaves is that both are used for _____.

Physics

2 answers:

Cloud [144]3 years ago

5 0

Answer:

(B) Sending Messages

Explanation:

A similarity between radio waves and microwaves is that both are used for sending messages and communication

Radio Waves and Microwaves have longer wavelengths and lower frequencies. Longer wavelengths than an x-ray which is in the electromagnetic spectrum and lower frequencies than gamma rays, which are also in the electromagnetic spectrum. They are also both electromagnetic radiation.

nikklg [1K]3 years ago

5 0

Sending messages because I said so and the guy on top said it so it’s right love u

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A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight

LuckyWell [14K]

Answer:

At an angle of $30^{\circ}$

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

$R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}$

6 0

3 years ago

an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati

kicyunya [14]

GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~

5 0

3 years ago

Read 2 more answers

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.

sergeinik [125]

Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²

5 0

3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

$v_2 = 5.6 \ m/s$

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$