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Andrews [41]
3 years ago
13

Complete this paragraph regarding dangers from the Sun.

Physics
2 answers:
Firlakuza [10]3 years ago
8 0

Answer:

[1]. b

[2].c

[3].a

[4].b

Explanation:

plato

i just took the test

Ganezh [65]3 years ago
4 0

Answer:

1: B. High-frequency

2:C. Satelites

3:A. Spacecraft

4:B. geomagnetic storms

i just took the test on PLATO

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When scientists work together it helps them achieve their goals when they share different ideas with eachother

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How many different types of atoms are present in one molecule of aluminum hydroxide, Al(OH)3?
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A man ran the 200m race within 19.19 seconds. Calculate the speed of the man during the race
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around 23 miles per hour

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The study and analysis of light according to its component wavelengths is called
BARSIC [14]

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The study and analysis of light according to its component wavelengths is called spectroscopy.

EXPLANATION:

Spectroscopy is the branch of science  that is concerned with the investigation and measurement of spectrum produced when matter interacts with or emits electromagnetic radiation.It helps us to identify atoms and molecules in the object.Spectroscopy is used to find out Dopplers effect (the red shift and blue shift),which tells how fast the object is comming towards earth or moving away from the earth.

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3 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
3 years ago
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