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DedPeter [7]
3 years ago
9

Identify the electrical properties of materials that are highly conductive and those that are used in insulation. Use one or two

sentences to complete your answer
Physics
1 answer:
Katena32 [7]3 years ago
6 0

Answer:

This question is incomplete

Explanation:

However, materials that are highly conductive have the capacity to allow heat and electricity to pass through them. This occurs because they have electrons that flow easily in them, these electrons serve as carriers of heat and electric current in these materials. It should be noted here that not only solid metals (like silver, aluminium and copper) are good conductors of electricity, liquid solution such as aqueous NaCl are also good conductors of electricity because of the same movement of electrons (which is being transferred during bonding as charges) in the liquid.

Invariably, materials that are poor conductors of heat and electricity do not have electrons moving freely in them. Examples are sand, rubber and plastic.

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An object carries a +15.5 uC charge.
abruzzese [7]

Answer:

3.67 N

Explanation:

From the question given above, the following data were obtained:

Charge of 1st object (q₁) = +15.5 μC

Charge of 2nd object (q₂) = –7.25 μC

Distance apart (r) = 0.525 m

Force (F) =?

Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:

For the 1st object

1 μC = 1×10¯⁶ C

Therefore,

15.5 μC = 15.5 × 1×10¯⁶

15.5 μC = 15.5×10¯⁶ C

For the 2nd object:

1 μC = 1×10¯⁶ C

Therefore,

–7.25 μC = –7.25 × 1×10¯⁶

–7.25 μC = –7.25×10¯⁶ C

Finally, we shall determine the force. This can be obtained as follow:

Charge of 1st object (q₁) = +15.5×10¯⁶ C

Charge of 2nd object (q₂) = –7.25×10¯⁶ C

Distance apart (r) = 0.525 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²

F = 3.67 N

Therefore, the force on the object is 3.67 N

4 0
3 years ago
an athlete at the Olympic Game run at constant speed and covers the 100 yard dash in 9.52seconds. what will be his time for 100
pickupchik [31]

Dr.phil can help you with that answer also the answer is 21.3

5 0
3 years ago
Explain how Law 1 applies to the image to the left.
alisha [4.7K]

Answer:

12

Explanation:

3 0
3 years ago
Read 2 more answers
A telephone line hangs between two poles 14 m apart in the shape of the catenary , where and are measured in meters.
Masja [62]

Answer:

a) At x=14 the slope will be given by:

\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right).

b) Then, the angle between the line and the pole will be:

\phi=\pi - \theta

where \theta is the angle between the tangent to the catenary and the x-axis.

Explanation:

The catenary has the following general form:

y(x)==a\cosh \left({\frac {x-C_{1}}{a}}\right)+C_{2}

a) The slope at any point will be given by the derivative of y.

\frac{dy}{dx}(x)=a\sinh \left({\frac {x-C_{1}}{a}}\right)

At x=14:

\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right).

b) The angle between the tangent to the catenary and the x-axis at a given point will be given by:

\frac{dy}{dx}(x)=tan(\theta) ⇒ \theta=tan^{-1} (\frac{dy}{dx}(x))

Then, the angle between the line and the pole will be:

\phi=\pi - \theta.

5 0
4 years ago
A 2kg object is dropped from a height 10m.Calculate the speed of the object after it has fallen 5m, assuming there is no resista
Ugo [173]

Answer:

10m/s

Explanation:

d=v_ot+\dfrac{1}{2}at^2

Since there is no initial velocity as the object is dropped, you can write the following equation:

5=\dfrac{1}{2}(10)t^2 \\\\1=t^2 \\\\t=1

Now that you know how long the fall took, you can use another physics equation to find the velocity at that point.

v_f=v_o+at

Since there once again is no initial velocity, you can rewrite this as:

v_f=at=(10)(1)=10m/s

Hope this helps!

5 0
4 years ago
Read 2 more answers
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