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lyudmila [28]
2 years ago
5

Explain how the Indian and Asian plates met and the kind of boundary that formed on their meeting. Describe what happened as a r

esult and tell why.
Physics
2 answers:
masya89 [10]2 years ago
8 0
225 million years ago (Ma) India was a large island situated off the Australian coast and separated from Asia by the Tethys Ocean. The supercontinent Pangea began to break up 200 Ma and India started a northward drift towards Asia. 80 Ma India was 6,400 km south of the Asian continent but moving towards it at a rate of between 9 and 16 cm per year. At this time Tethys Ocean floor would have been subducting northwards beneath Asia and the plate margin would have been a Convergent oceanic-continental one just like the Andes today.
ANEK [815]2 years ago
4 0

Answer:

This immense mountain range began to form between 40 and 50 million years ago, when two large landmasses, India and Eurasia, driven by plate movement, collided. ... Artist's conception of the 6,000-km-plus northward journey of the "India" landmass (Indian Plate) before its collision with Asia (Eurasian Plate).

Explanation:

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Y is a proton and z is an electron. Protons and neutrons are in the nucleus.
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An LED with total power P tot = 960 mW emits UV light of wavelength 360 nm. Assuming the LED is 55% efficient and acts as an iso
Anit [1.1K]

Answer:

E_0=225.09N/C

Explanation:

We are given that

Power,Ptot=960mW=960\times 10^{-3}W

1mW=10^{-3} W

Wavelength,\lambda=360 nm=360\times 10^{-9} m

1nm=10^{-9} m

Distance,r=2.5 cm=2.5\times 10^{-2} m

1m=100 cm

Efficiency=55%

Power radiation emitted=\frac{55}{100}\times 960\times 10^{-3}=0.528W

Intensity,I=\frac{P}{4\pi r^2}

I=\frac{0.528}{4\pi(2.5\times 10^{-2})^2}=67.26W/m^2

Intensity,I=\frac{1}{2}c\epsilon_0E^2_0

E^2_0=\frac{2I}{c\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}

c=3\times 10^8 m/s

E_0=\sqrt{\frac{2I}{c\epsilon_0}}

E_0=\sqrt{\frac{2\times 67.26}{3\times 10^8\times 8.85\times 10^{-12}}}

E_0=225.09N/C

4 0
2 years ago
A man stands on a scale and holds a heavy object in his hands. What happens to the scale reading if the man quickly lifts the ob
Novosadov [1.4K]

Answer:

Explanation:

When he accelerates the heavy object up , the reading increases because an extra downward normal force acts on it, then scale reading returns to the same reading as when standing stationary, and then decreases as although he is lifting the heavy object , the acceleration is decreasing ,so the extra upward normal force acts.

6 0
3 years ago
Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge
butalik [34]

Answer:

Electric charge in the earth will be Q=5.231\times 10^5C

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}. here K is coulomb's constant

So  116=\frac{9\times 10^9\times Q}{(6371000)^2}

Q=5.231\times 10^5C

So electric charge in the earth will be Q=5.231\times 10^5C

5 0
2 years ago
How high can a body vertically thrown with a speed of 40m/s raise after 3 sec (neglecting air
Tcecarenko [31]

y = 75.9 m

Explanation:

y = -(1/2)gt^2 + v0yt + y0

If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.

y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)

= -44.1 m + 120 m

= 75.9

5 0
2 years ago
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