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2 years ago

PLEASE HELP ASAP! Any absurd answers will be reported. NO LINKS, IDC IF ITS THE ONLY WAY TO GET AN ANSWER.

Physics

2 answers:

maxonik [38]2 years ago

8 0

Www.cheese.com

Report me and I’ll report u for harassment:)

svetoff [14.1K]2 years ago

6 0

Answer:

There is a thing called a continental drift. It started about 200 million years ago. At first the continents were all attached, this super continent was called pangaea. Continental drift occurs because of the shift of the tectonic plates within the earth's outer shell. The heat from within the earth triggers movement to occur. This a very slow process though. It took 200 million years for the continents to get where the are now and would probably take another 200 to collide.

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A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh

Artyom0805 [142]

Answer:

The electric field is $5.623\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 0.357 m

Charge $Q=5.03\times10^{-6}\ C$

Point charge $q=4.15\times10^{-6}\ C$

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}$

Where, q = point charge

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}$

$E=5.623\times10^{4}\ N/C$

Hence, The electric field is $5.623\times10^{4}\ N/C$

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2 years ago

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

Westkost [7]

Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

To calculate the time the object takes to hit the ground, we use the equation below:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

$\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01$

d = 15.15 m

The time in the air is:

$\displaystyle t=\sqrt{\frac{2*5}{9.8}}$

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

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2 years ago

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Answer:

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Answer:

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