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4 years ago

Describe two examples of how humans use electromagnetic waves.

2 answers:

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]4 years ago

4 0

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

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Samaira needs to rent some tents for an outdoor family reunion in July what is the best type of tent for Samaira to rent so that

Elena L [17]

Answer:

It seems that you have missed the necessary options for us to answer this question, so I had to look for it. Anyway, here is the answer. Samaira needs to rent some tents for an outdoor family reunion in July, so the type of tent for Samaira to rent so that her family members are protected from the heat of the sun is SMOOTH and WHITE TENT. Hope this answer helps.

6 0

3 years ago

Read 2 more answers

A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?

Vinil7 [7]

Answer:

5010J

Explanation:

Work is force times distance, or

W=F⋅d.

Substitute in values from the question to get

W=8.35⋅102N⋅6m=50.1⋅102Nm=5010J

6 0

3 years ago

Read 2 more answers

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

Polar moment of Inertia

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

Maximum sustainable torque on the solid circular shaft

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

Maximum sustainable torque on the tubular shaft

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

Maximum sustainable power in the solid circular shaft

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

Maximum sustainable power in the tubular shaft

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

An x-ray has a wavelength of 4.18 Å. Calculate the energy (in J) of one photon of this radiation. Enter your answer in scientifi

Damm [24]

Answer:

E = 4.75 x 10⁻¹⁶ J

Explanation:

given,

wavelength of the x-ray , λ = 4.18 Å

Energy of photon = ?

we know

$E = \dfrac{hc}{\lambda}$

where h is the planks constant

c is the speed of light

h = 6.626 x 10⁻³⁴ m² kg / s

c = 3 x 10⁸ m/s

now,

$E = \dfrac{6.626\times 10^{-34}\times 3 \times 10^8}{4.18\times 10^{-10}}$

E = 4.75 x 10⁻¹⁶ J

hence, the energy of the photon is equal to E = 4.75 x 10⁻¹⁶ J

4 0

4 years ago

Many physical quantities are connected by inverse square laws, that is, by power functions of the form f(x)=kx^(-2). In particul

goldenfox [79]

Answer:

4 times

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is inversely proportional to the square of the distance from the light source, the form of the function is:

f(x) = k.x⁻²

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

$F=k(x/2)^{-2}=\frac{k}{(x/2)^2}= \frac{k}{x^2}. 4=f(x).4$

Then, you have found that the light is 4 times as bright as it originally was.

3 0

3 years ago