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JulijaS [17]
3 years ago
9

Which of the following is an ionic compound?

Physics
1 answer:
const2013 [10]3 years ago
7 0

compound is covalent! The bond between HI is a polar covalent bond. ... It is because Hydrogen is electropositive in nature in comparison to Iodine.

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A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance
yan [13]

Answer:

a)  h'= 5/7 h , b)   h ’= h

Explanation:

Let's use energy conservation for this exercise

Starting point. Upper left side

         Em₀ = mg h

Final point. Lower left side

        Emf = K = ½ m v² + ½ I w²

        Em₀ = emf

        mgh = ½ m v² + ½ I w²

Angular and linear velocity are related

           v = r w

           w = v / r

The moment of inertia of the marble that we take as a solid sphere is

           I = 2/5 m r²

We substitute

          m g h = ½ m v² + ½ 2/5 m r² (v / r)²

          g h = ½ v2 (1 + 2/5)

         v = √(g h 10/7)

This is the speed at the bottom of the bowl

Now let's apply energy conservation to the right side

a) right side if rubbing

             Em₀ = K

              Emf = U = mg h’

             ½ m v² = mg h’

              h’= ½ (g h 10/7) / g

              h'= 5/7 h

b) right side with rubbing

             Em₀ = K

             Emf = K + U = -½ I w² + m g h

             Emf = -½ 2/5 m r² v² / r² + m gh

            Em₀ = emf

            ½ v² = -1/5 v² + g h’

            h’= (1/2 +1/5) (gh 10/7) / g

            h ’= h

c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher

8 0
3 years ago
You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in
hoa [83]

Answer:F/L = μ0*I²/2πd

Explanation:

Both wires carries current of the value (I) and are separated by the distance (d).

The length of both wires is (L).

The first wire creates a magnetic field of the magnitude according to Bio-savart law

B=(Uo * I)/2πd

This magnetic field has the ability to exert a force F on the second wire.

This force (F) is given as

F=BIL* sin θ

θ here is 90 because the magnetic field is perpendicular to the length of the second wire.

Thus the value of sin 90 = 1

Hence F= BIL

But B=(Uo * I)/2πd.

We have that F= B=[(Uo * I)/2πd] x IL

Thus we have that F= [Uo* I²/2πd] x L

By rearranging, and bringing L to the left hand side of the equation, we have that

F/L = [Uo* I²/2πd]

6 0
3 years ago
Your mobile charger is an example of<br>a.transformer<br>b.solenoid<br>c.motor.<br>d.generator<br>​
lianna [129]

Answer:

Your mobile charger is an example of (d.) a generator.

Explanation:

Your mobile charger generates energy to charge your device.

5 0
3 years ago
Read 2 more answers
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
irinina [24]

Answer:

a) \sigma_{\rm in} = -2.18~{\rm \mu C/m^2}

b) \sigma_{\rm out}= 1.12~{\rm \mu C/m^2}

c) E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}

Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}

c) The electric field outside the cylinder can be found by Gauss' Law:

\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

4 0
3 years ago
Identify the following as
bazaltina [42]
3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy
6 0
3 years ago
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