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vivado [14]
3 years ago
10

Which of the following statements is the most accurate about the Earth’s average temperature over the last 30 years?

Physics
1 answer:
enyata [817]3 years ago
5 0

Answer:

the 3rd one

Explanation:

climate change

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Answer:

BET SHORTY

Explanation:

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bcz value of gravatation on jupiter is high in comparison with earth and it will keep hard to jump on jupiter

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3 years ago
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When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0
3 years ago
A delusion is best characterized as
balu736 [363]

Answer:

This is the most common form of delusional disorder. In this form, the affected person fears they are being stalked, spied upon, obstructed, poisoned, conspired against or harassed by other individuals or an organization.

Explanation:

hope this help

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3 years ago
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Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days.
kvasek [131]

Answer:

Its rotation will be 3.89x10⁴ rad/s.

Explanation:

We can find the rotation speed by conservation of the angular momentum:

L_{i} = L_{f}

I_{i}\omega_{i} = I_{f}\omega_{f}   (1)

The initial angular speed is:

\omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d}

The moment of inertia (I) of a sphere is:

I = \frac{2}{5}mr^{2}    (2)

Where m is 9 times the sun's mass and r is the sun's radius

By entering equation (2) into (1) we have:

\frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f}

9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}

\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s          

Hence, its rotation will be 3.89x10⁴ rad/s.

I hope it helps you!                                                        

3 0
4 years ago
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