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marusya05 [52]
3 years ago
15

You are going to invent a new device that people can use instead of a microwave oven to warm up leftovers. Which type of electro

magnetic wave would be most useful to investigate?
gamma rays
X–ray
radio waves
infrared waves
Physics
2 answers:
kicyunya [14]3 years ago
7 0
If y<span>ou are going to invent a new device that people can use instead of a microwave oven to warm up leftovers, the type of electromagnetic wave that would be most useful to investigate would probably be infrared waves, because these kinds of waves are associated with heat. </span>
77julia77 [94]3 years ago
4 0

Answer:

Infrared Waves

Explanation:

Since electromagnetic waves are of different energy range is used for different works

Higher energy range electromagnetic waves are not good to expose for human so we can not use them for warming purpose as they are dangerous.

So X rays and Gamma rays can not be used

Also radio waves are of least energy so they can not be used for the purpose of warming.

So correct answer for the replacement of microwaves is infrared waves because they have sufficient energy that can be used to warm up.

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An LED with total power P tot = 960 mW emits UV light of wavelength 360 nm. Assuming the LED is 55% efficient and acts as an iso
Anit [1.1K]

Answer:

E_0=225.09N/C

Explanation:

We are given that

Power,Ptot=960mW=960\times 10^{-3}W

1mW=10^{-3} W

Wavelength,\lambda=360 nm=360\times 10^{-9} m

1nm=10^{-9} m

Distance,r=2.5 cm=2.5\times 10^{-2} m

1m=100 cm

Efficiency=55%

Power radiation emitted=\frac{55}{100}\times 960\times 10^{-3}=0.528W

Intensity,I=\frac{P}{4\pi r^2}

I=\frac{0.528}{4\pi(2.5\times 10^{-2})^2}=67.26W/m^2

Intensity,I=\frac{1}{2}c\epsilon_0E^2_0

E^2_0=\frac{2I}{c\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}

c=3\times 10^8 m/s

E_0=\sqrt{\frac{2I}{c\epsilon_0}}

E_0=\sqrt{\frac{2\times 67.26}{3\times 10^8\times 8.85\times 10^{-12}}}

E_0=225.09N/C

4 0
3 years ago
If a system requires 150 j of input work and produces 123 J of output work, whats its effiency
Yuki888 [10]

The effiency of a machine is

                 (output work or energy) / (input work or energy) .

For the system described in the question, that's

                 (123 J) / (150 J) = 0.82  =  82% .

 
4 0
3 years ago
how fast in revolutions per minute must a centrifuge rotate in order to subject the contents of a test tube (30cm) from the axis
trapecia [35]

Answer:

ω=6684.51 rpm

Explanation:

r= 30cm= 0.3m

a= 15000gs (convert to m/s^{2}

1g = 9.8 m/s^{2}

a= 15000 *9.8 = 147000 m/s^{2}

a=\frac{v^{2} }{r}

147000 = \frac{v^{2} }{0.3}

147000*0.3 = v^{2}

44100 = v^{2}

√44100 = v

210m/s  = v

v=210m/s (linear velocity)

we will convert this to angular velocity

ω=\frac{v}{r}  

ω= 210/0.3

ω= 700 rads^{-1}

we will convert this to rev per minute

1rad per second = 9.5493 rev per minute

ω= 700*9.5493

ω=6684.51 rpm

4 0
4 years ago
Tyson Gay's best time to run 100.0 meters was 9.69 seconds. What was his average speed during this run, in miles per hour? (3.28
FinnZ [79.3K]

Answer:

23.086 mile/h

Explanation:

Given,

Distance Tyson Gay run = 100 m

time of run, t = 9.69 s

average speed of the in mph = ?

Speed of the Gay = \dfrac{distance}{time}

v = \dfrac{100}{9.69}

     v = 10.32 m/s

1 m = 3.281 ft

10.32 m = 33.86 ft

1 mile = 5280 ft

1 ft = 1.8939 x 10⁻⁴ mile

33.86 ft/s = 6.413 x 10⁻³ miles/s

Speed of Tyson in mile/hr = 6.413 x 10⁻³ x 3600

                                           = 23.086 mile/h

Hence, speed of Tyson Gay's in mile/ hr is equal to 23.086 mph.

4 0
3 years ago
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What is a analogy for transition metals
Ganezh [65]

<em>A simple metallic band model is proposed for the transition metal mono antimonides, by analogy to the transition metals.</em>

6 0
3 years ago
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