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Alexxandr [17]
3 years ago
14

Why would scientist use both relative and absolute dating

Physics
1 answer:
Gemiola [76]3 years ago
6 0
Well, relative dating is the science determining the relative order of past events like the age of an object compared to another object without determining the absolute age. The absolute age is the process of determining an age on a specific chronology in archeology and geology. Some scientists prefer the terms chromatic or calendar dating, as the word absolute implies an unwarranted certainty or accuracy. Hope I helped
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A plane flying at a steady speed of 100 m/s accelerates to 150 m/s in 10 seconds. What is the plane’s acceleration?
Rashid [163]

A plane flying initially at 100 m/s uses an acceleration of 5 m/s² to reach a velocity of 150 m/s in 10 seconds.

<h3>What is acceleration?</h3>

Acceleration is the change in velocity over time.

A plane is flying initially at 100 m/s (u) and it accelerates to 150 m/s (v) in 10 s (t). We can calculate its acceleration using the following expression.

a = v - u / t = (150 m/s - 100 m/s) / 10 s = 5 m/s²

A plane flying initially at 100 m/s uses an acceleration of 5 m/s² to reach a velocity of 150 m/s in 10 seconds.

Learn more about acceleration here: brainly.com/question/14344386

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5 0
1 year ago
What is the current in a 160V circuit if the resistance is 20Ω
Mkey [24]

Answer:

I =V/R

I= 160/20 = 8 Amperes

3 0
3 years ago
You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in
ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f  = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5  + 0.6 × 20c = 343.5

so wavelength of sound  λ = c / f  

wavelength = 343.5 /  686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............   for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475  = 2.975 m

distance 2 = 2.5 + 3.125  = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0
3 years ago
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