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3 years ago

Which layer of the atmosphere has the highest density?

Physics

1 answer:

fomenos3 years ago

4 0

Answer:

The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.

Explanation:

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It's Endorphins. That's a pain killer produced by the brain.

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3 years ago

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3 years ago

Which contribution did Kepler make to viewing the solar system?

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3 years ago

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,

KATRIN_1 [288]

Answer:

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

V = i (R₁ + R₂)

with R₁ connected

V = (i + 0.5) R₁

with R₂ connected

V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) = $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

i² + 0.75 i + 0.125 = 2i² + 0.75 i

i² - 0.125 = 0

i = √0.125

i = 0.35355 A

with the second equation we look for R1

R₁ = $\frac{V}{i+0.5}$

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ = $\frac{V}{i+0.25}$

R₂ = $\frac{12}{0.35355+0.25}$

R₂ = 19.9 Ω

3 0

3 years ago

A car moving at 10 m/s crashes into a large bush and stops in 1.3 m. Using the Work-Energy theorem, calculate the average force

ad-work [718]

Answer:

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

It takes 0.26 s to bring the passenger in the car to a halt.

Explanation:

Hi there!

The negative work (W) needed to bring a moving object to stop is equal to its kinetic energy (KE):

W = KE

F · s = 1/2 · m · v²

Where:

F = applied force on the passenger.

s = displacement of the passenger.

m = mass of the passenger.

v = velocity of the passenger.

Solving the equation for F:

F = 1/2 · m · v² / s

Replacing with the data:

F = 1/2 · 70 kg · (10 m/s)² / 1.3 m

F = 2692.3 N

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

According to the second law of Newton:

F = m · a

Where "a" is the acceleration of the passenger.

We also know from kinematics that the velocity of an object can ve calculated as follows:

v = v0 + a · t

Where:

v = velocity of the passenger at time t.

v0 = initial velocity.

t = time.

a = acceleration.

When the passenger stops, its velocity is zero. So replacing a = F/m, let´s solve the equation for the time it takes the passenger to stop:

v = v0 + a · t

0 = 10 m/s + (-2692.3 N/ 70 kg) · t

-10 m/s / (-2692.3 N/ 70 kg) = t

t =0.26 s

It takes 0.26 s to bring the passenger in the car to a halt.

5 0

3 years ago