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Alex777 [14]
3 years ago
10

Two 2.0-cm-diameter insulating spheres have a 6.60 cm space between them. One sphere is charged to + 76.0 nC , the other to - 30

.0 nC . Part A What is the electric field strength at the midpoint between the two spheres? Express your answer with the appropriate units.
Physics
1 answer:
e-lub [12.9K]3 years ago
8 0

Answer:

5.2\times 10^5N/C

Explanation:

Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.

This can be easily seen if we use Gauss's law, \int{E} \, dA=\frac{Q_{enclosed}}{\epsilon_o}

We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,

E=\frac{1}{4\pi r^2} \frac{Q_{enclosed}}{\epsilon_o}

which is the same as that of a point charge.

In our problem the charges being of opposite signs, the electric field will add up. Therefore,

E_{total}=\frac{1}{4\pi\epsilon_o}\frac{q_1+q_2}{r^2}= (9\times10^9) \frac{(76+30)\times10^{-9}}{((1+3.3)\times10^{-2})^2}N/C =5.2\times10^5N/C

where, r = distance between the center of one sphere to the midpoint (between the 2 spheres)

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Which element is located in period 3 of the periodic table?
liberstina [14]

Answer:

From the given options, the option (D) is correct. Because Al is the aluminium which is located in period 3 of the periodic table.

Explanation:

We know that the 3rd period contains 8 elements which are as follows:

  1. sodium
  2. magnesium
  3. aluminium
  4. silicon
  5. phosphorus
  6. sulfur
  7. chlorine, and
  8. argon

The sodium and magnesium, the first 2 elements, are members of the s-block of the periodic table, while the rest of the members are the elements of the p-block.

Thus, from the given options, the option (D) is correct. Because Al is the aluminium which is located in period 3 of the periodic table.

3 0
2 years ago
A rectangular block has the density of 350g/cm3 the dimensions are 3.5 6.cm 2.5
LekaFEV [45]

Answer:

18375g

Explanation:

\boxed{density =  \frac{mass}{volume} }

∴ mass = density \times volume

Let's find the volume of the rectangular block.

Volume

= length ×breadth ×height

= 3.5 ×6 ×2.5

= 52.5cm³

Mass of the block

= 350(52.5)

= 18375g

7 0
2 years ago
The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation b
d1i1m1o1n [39]
The electrostatic force between two charges Q1 and q is given by
F=k_e  \frac{Q_1 q}{r^2}
where 
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges

Re-arranging the formula, we have
q= \frac{F r^2}{k_e Q_1}
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
q= \frac{(0.961 N)(0.67 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.44\cdot 10^{-6}C)}=5.69 \cdot 10^{-6} C

And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.
6 0
3 years ago
Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that
Goryan [66]

Answer:

Explanation:

In the decibel scale , intensity of sound changes logarithmically as follows

10log\frac{I}{I_0} = Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

10log\frac{I}{10^{-12}} = 71

log\frac{I}{10^{-12}} = 7.1

\frac{I}{10^{-12}} = 10^{7.1}

I= 10^{-4.9} W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x 10^{5.4}

I= 10^{-6.6 } W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

I= 10^{-4.9}  = 2π² x 504²xA²x 1.21 x  346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

10^{-6.6} = 2π² x 504²xA²x 1.21 x  346

A² = 1.1978 x 10⁻¹⁶

A = 1.1 x 10⁻⁸ m

6 0
3 years ago
A.) Determine the work done by Zach on the bull.
Lana71 [14]

Explanation:

Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

    Work done  = Force x distance

The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

 So to solve this problem;

 multiply the force applied by Zack and distance through which the bull was pulled.

3 0
2 years ago
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