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murzikaleks [220]

3 years ago

7

Describe two examples of how humans use electromagnetic waves.

2 answers:

Alex73 [517]3 years ago

5 0

Electromagnetic waves are used in everyday life. You are looking at your computer screen right now. The light that is coming off of the screen is visible light, a form of electromagnetic radiation. Electromagnetic waves are also used to send information. For example, AM or FM radios are radio waves that transfer sound information to your local radio.

cricket20 [7]3 years ago

4 0

We used electromagnetic waves in Cell Phones and Radios as well as Microwaves!! Hope that helps!!

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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

nalin [4]

Answer:

$13 W/m^2$

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

$I \propto \frac{1}{r^2}$

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$ (1)

where in this problem, we have:

$I_1 = 1300 W/m^2$ apparent brightness at a distance $r_1$ , where

$r_1 = 150$ million km

We want to estimate the apparent brightness at $r_2$ , where $r_2$ is ten times $r_1$ , so

$r_2 = 10 r_1$

Re-arranging eq.(1), we find $I_2$ :

$I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2$

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2 years ago

Which action will increase the gravitational force between two objects?

schepotkina [342]

The answer is D have a nice day!

8 0

2 years ago

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Using poiseulli assumption derive an expression for the rate of flow of a liquid through a horizontal tube in terms of a, l, p a

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Answer:

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3 years ago

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

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3 years ago

Suppose a ball of putty moving horizontally with 1kg.m/s of momentum collides and sticks to an identical ball of puffy moving ve

WARRIOR [948]

Answer:

Vector sum of two vectors at right angles

p={p₁²+p₂²} =2 =1.41 kg•m/s

Explanation:

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3 years ago