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slamgirl [31]
2 years ago
11

There is up to 30 times more gold in a tons of old mobile phones than in a tons of gold ore. true or false

Physics
1 answer:
FrozenT [24]2 years ago
3 0

Answer:

true

Explanation:

dbgkdodocofkci ifkdcl k kfododocp v

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What happens when the objects submerged in a fluid at rest?​
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It will act upon a buoyant force on the magnitude of which is equal to weight of the fluid
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A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is th
Genrish500 [490]

Answer:

L^2(\dfrac{8m}{3}+m_r)

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

m_r = Mass of ring

Moment of inertia of a spoke

\dfrac{mL^2}{3}

For 8 spokes

8\dfrac{mL^2}{3}

Moment of inertia of ring

m_rL^2

Total moment of inertia

8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is L^2(\dfrac{8m}{3}+m_r).

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3 years ago
What is larger a Atom or a Molecule?
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I believe an Atom is a very powerful source, the basic unit of a chemical element. An atom is a source of nuclear energy.


But a molecule on the other hand isn't so different.
a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
I hope that helps, have a fantastic day!
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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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