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3 years ago

what would happen to the surface temperature on Earth if large amounts of carbon dioxide were removed from the atmosphere?

Physics

1 answer:

slava [35]3 years ago

8 0

Answer:

If carbon emissions stopped, the oceans catch up with the atmosphere, the Earth's temperature would rise about another 1.1F (0.6C).

Explanation:

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A body of mass 2 kilograms moves on a circle of radius 3 meters, making one revolution every 5 seconds. Find the magnitude of th

harkovskaia [24]

Answer: $F_c=9.375 N$

Explanation:

Given

mass of body $m=2 kg$

radius of circle $r=3 m$

Time Period $T=5 s$

suppose $\omega$ is the angular velocity of revolution

therefore $\omega T=2\pi$

$\omega =\frac{2\pi }{5}=1.25 rad/s$

Centripetal acceleration $a_c=\omega ^r$

$a_c=1.25^2\times 3$

$a_c=4.68 m/s^2$

therefore centripetal $F_c=m\omega ^2\times r$

$F_c=9.375 N$

7 0

3 years ago

Can someone help me with these two 2 question 3 and 4 please

9966 [12]

3: C the earth's axis is angled meaning the sun is closer or farther to/from the earth during season changes
4: False a revolution is how long the earth takes around the sun

6 0

3 years ago

How long will it take you to drive the 256 miles to get to the Mackinac Island Ferry if you average 70 mi/hr? If you leave at 3:

11111nata11111 [884]

Answer:

If the speed is 70 mile/hour, the time to drive the 256 miles is:

T = 256/70 = 3.66 hours = 3 hours and 39.6 minutes

If you leave at 3:00, then at 6 hours and 39.6 minutes, you will get there.

Hope this helps!

8 0

4 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

4 0

3 years ago

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

Since the speed of sound is not given in the question, we would assume it to be 340m/s

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

So, the length of the second pipe is L – L’

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

The beat frequency due to the fundamental frequencies of the first and second pipe is

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago