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3 years ago

What are the upper boundaries of each layer of the atmosphere

1 answer:

6 0

The atmosphere is composed of 5 layers which is 500 kilometers above the ground. This are Troposphere, Stratosphere, Mesoshpere, Thermosphere, and Exosphere. After the Exosphere is the outer bounderies which is the space

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A 2.3 kg block executes SHM while attached to a horizontal spring of spring constant 150 N/m. The maximum speed of the block as

aliya0001 [1]

Answer

given,

mass of block = m = 2.3 Kg

spring constant = k = 150 N/m

speed = 2.3 m/s

a) we know,

$\omega = \sqrt{\dfrac{k}{m}}$

$\omega = \sqrt{\dfrac{150}{2.3}}$

ω = 8.08 rad/s

v = A ω

2.3 = A x 8.08

A = 0.285 m

b) maximum acceleration

a = A ω²

a = 0.285 x 8.08²

a = 18.58 m/s²

c) acceleration is minimum at mean position the minimum value of acceleration is zero.

d) time period

$T = \dfrac{2\pi}{\omega}$

$T = \dfrac{2\pi}{8.08}$

T = 0.778 s

t = 4.1 x 0.7778

t = 3.188 s

3 0

4 years ago

What a scientist often conducts a controlled experiment,which is?

Wittaler [7]

8 the height variations

where the water is

7 0

3 years ago

Read 2 more answers

Rahul sees a flock of birds. He watches as the flying birds land in neat little rows on several power lines. Which change of sta

Murljashka [212]

The change of state does the motion of the birds most closely model is deposition. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

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3 years ago

Read 2 more answers

A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the

HACTEHA [7]

The work done onto the car is 506,250 J

The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.

From the parameters given:

The mass of the car = 1500 kg
The initial speed = 30 m/s
The final speed = 15 m/s

The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)

$\mathbf{=\dfrac{1}{2} mv_1^2 -\dfrac{1}{2} mv_2^2}$

$\mathbf{=\dfrac{1}{2} m(v_1^2 - v_2^2)}$

$\mathbf{=\dfrac{1}{2} \times 1500 \times (30^2 - 15^2)}$

= 506,250 J

Therefore, we can conclude that the work done on the car is 506,250 J

Learn more about work done here:

brainly.com/question/18762601

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3 years ago

A beam of light traveling in air strikes a glass slab at an angle of incidence less than 90°. After entering the glass slab, wha

ad-work [718]

Answer:

True A and B

Explanation:

Let's propose the solution of the exercise before seeing the affirmations.

We use the law of refraction

n₁ sin θ₁ = n₂. Sin θ₂

Where n₁ and n₂ are the refractive indices of the two means, θ₁ and θ₂ are the angles of incidence and refraction, respectively

sin θ₂ = (n1 / n2) sin θ₁

Let's apply this equation to the case presented. The index of refraction and airs is 1 (n1 = 1)

Sin θ₂ = (1 / n2) sin θ₁

the angle θ₂ which is the refracted angle is less than the incident angle

Let's analyze the statements time

A. False. We saw that it deviates

B. True Approaches normal (vertical axis)

C. False It deviates, but it is not parallel to normal

D. False It deviates, but approaching the normal not moving away

E. True. Because its refractive index is higher than air,

8 0

3 years ago