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juin [17]
4 years ago
14

Match the EM wave type to the correct description. A. Gamma ray B. X-ray C. Ultraviolet D. Visible light E. Infrared F. Microwav

e G. Radio/TV waves - Part of the spectrum the human eye can see- Highest energy EM wave-High frequency wave dangerous to human skin-Often used for medical imaging-Medium frequency wave-Low energy wave sometimes used to heat food-Low energy wave sometimes used to heat food-Longest wavelength EM wave
Physics
2 answers:
Blababa [14]4 years ago
6 0
This is the order of the spectrum from highest frequency to lowest frequency:
Gamma Ray (Highest energy em wave)
X Ray (Used in medical imaging)
Ultra violet (dangerous to human skin)
visible light (human eye can see)
Infrared (medium frequency)
Microwave (used to heat food)
Radio (longest wavelength)

Flura [38]4 years ago
6 0

Answer:

A. Gamma ray

- Highest energy EM wave

B. X-ray

-Often used for medical imaging

C. Ultraviolet

-High frequency wave dangerous to human skin

D. Visible light

- Part of the spectrum the human eye can see

E. Infrared

-Medium frequency wave

F. Microwave

-Low energy wave sometimes used to heat food-Low energy wave sometimes used to heat food

G. Radio/TV waves

-Longest wavelength EM wave

Explanation:

As we arrange the electromagnetic waves in increasing order of energy and frequency from lowest to highest then it is given as

1. Radio Waves

2. Microwaves

3. Infrared waves

4. Visible Light

5. Ultraviolet waves

6. X - Rays

7. Radio waves

So as per the energy range we can say most energetic waves are gamma rays and least energy waves are Radio waves and their uses are as mentioned above

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When atoms are split, they release energy. This concept applies to (2 points)
cupoosta [38]

This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.

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│Hope this helped  _____________________│    

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6 0
3 years ago
Its mass is 20 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume?
Alik [6]
Volume = mass / density
Volume = 20 / 7.87
Volume = 2.54 (2 s.f)
3 0
3 years ago
A proton with mass 1.67*10^-27kg is propelled at an initialspeed of 3.00*10^5m/s directly toward a uranium nucleus 5.00maway. Th
Tresset [83]

Answer:

Explanation:

F = 2.12 x 10⁻²⁶ / x²

Work done by electric field of nucleus

W = ∫ Fdx

= ∫2.12 x 10⁻²⁶ / x² dx

= 2.12 x 10⁻²⁶ ( - 1 / x )

= - 2.12 x 10⁻²⁶ ( 1/5 - 1 / 8 x 10⁻¹⁰ )

= - .265 x 10⁻¹⁶ J

1/ 2 x mv² = .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ - .265 x 10⁻¹⁶

= 7.515 x 10⁻¹⁷ - .265 x 10⁻¹⁶

=( .7515 - .265 )x 10⁻¹⁶

= .4865 x 10⁻¹⁶

.5 x 1.67 x 10⁻²⁷ x v² = .4865 x 10⁻¹⁶

v² = .5826 x 10¹¹

v² = 5.826 x 10¹⁰

v = 2.41 x 10⁵ m /s

b )

Let r be the closest distance

Potential at this point

2.12 x 10⁻²⁶ (  1 / r )

Kinetic energy

= 0

Total energy = 2.12 x 10⁻²⁶ (  1 / r )

Total energy at 5 m

= .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ + 0 ( potential energy at 5 m will be negligible as compared with that near the center )

= 7.515 x 10⁻¹⁷ J

So ,

2.12 x 10⁻²⁶ (  1 / r ) = 7.515 x 10⁻¹⁷

r = 2.12 x 10⁻²⁶ / 7.515 x 10⁻¹⁷

= .282 x 10⁻⁹

= 2.82 x 10⁻¹⁰ m

c ) As electric field is conservative , no dissipation of energy takes place . Hence its speed at 5m on returning back to this point will be same as

3.00 x 10⁵ m /s

7 0
4 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
Describe how the resistance of the filament lamp changes as the current through it increases.
pochemuha
This heats up when an electric current passes through it, and produces light as a result. The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.
8 0
3 years ago
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