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alex41 [277]
3 years ago
13

Name at least 4 other gases in the atmosphere besides oxygen and nitrogen

Physics
2 answers:
lukranit [14]3 years ago
4 0
Carbon dioxide,hydrogen,argon,helium
Irina18 [472]3 years ago
3 0
Carbon dioxide
Helium
Argon
Hydrogen
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Which of the following best defines amplitude?
ziro4ka [17]
The answer to that question is C
8 0
3 years ago
A car accelerates at a constant rate of 3 m/s2 for 5 seconds. If it reaches a velocity of 27 m/s, what was its initial velocity?
Kaylis [27]

The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

CALCULATE INITIAL VELOCITY:

The initial velocity of the car can be calculated by using one of the equation of motion as follows:

V = u + at

Where;

  • V = final velocity (m/s)
  • u = initial velocity (m/s)
  • a = acceleration due to gravity (m/s²)
  • t = time (s)

According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:

u = v - at

u = 27 - 3(5)

u = 27 - 15

u = 12m/s.

Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

Learn more about motion at: brainly.com/question/974124

4 0
2 years ago
A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0
2 years ago
Just wondering if I did this right
Kisachek [45]

Yeah

All they are all correct

5 0
3 years ago
Can someone help me with a)​
Gnom [1K]

Answer:

what's your problem. How may I help you Buddy

3 0
3 years ago
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