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Rus_ich [418]
2 years ago
8

Help please help!!!!!!!!

Physics
1 answer:
Sidana [21]2 years ago
6 0

1. Electron

2. Neutron

3. Nucleus

4. Atom

5. Molecule

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Read 2 more answers
This problem has been solved!
lana66690 [7]

Answer:

Charge on each metal sphere will be 8\times 10^{8}C

Explanation:

We have given number of electron added to metal sphere A n=10^{12}electron

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres =\frac{10^{12}}{2}=5\times 10^{11}electron

We know that charge on each electron e=1.6\times 10^{-19}C

So charge on both the spheres will be equal to q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C

So charge on each metal sphere will be equal to 8\times 10^{8}C

6 0
2 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
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