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3 years ago

How does temperature change in layers of the atmosphere?

Physics

1 answer:

Gemiola [76]3 years ago

4 0

Answer:

Explanation:

The troposphere is hotter near the Earth's surface because heat from the Earth warms this air. ... Mesosphere: As the altitude increases, the air temperature decreases. The Mesosphere, like the troposphere layer, has a decrease in temperature with altitude because of the decreases in the density of the air molecules.

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Light can make things happen when it is ______. Fill in the missing word. COrrect answer gets brainliest!

MatroZZZ [7]

Answer:

Light can make things happen when it is shining.

(or glowing)

8 0

3 years ago

Read 2 more answers

You need to know the height of a tower, butdarkness obscures the ceiling. You note thata pendulum extending from the ceiling alm

AysviL [449]

Answer:

The towers height is 322m

Explanation:

We can assume that a simple pendulum hangs from the top of the tower.

In this case, the length of the pendulum will be the same as the height of the tower.

We will use the next expression then:

$T=(2pi)(\sqrt{L/g})$

Where T is the period, L is the pendulum length and g is the gravity $(9.81m/s^2)$

From the previous expression we will have then that the length will be:

$L=(g)*((T/2pi)^2)$

Replacing:

$L=(9.81m/s^2)*((36s/2pi)^2)$

The towers height will be then: 322m

8 0

3 years ago

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi

vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel $N_1=2\ rev/s$

angular speed $\omega_1=2\pi N_1=4\pi\ rad/s$

mass of wheel $m_1=4.5\ kg$

diameter of wheel $d_1=0.30\ m=30\ cm$

radius of wheel $r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm$

mass of clay $m_2=2.8\ kg$

the radius of the chunk of clay $r_2=8\ cm$

Moment of inertia of Wheel

$I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2$

Combined moment of inertia of wheel and clay chunk

$I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2$

Conserving angular momentum

$\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi$