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nadezda [96]
2 years ago
6

Which of the following is a good insulator? Wire, doorknob, plastic, nickle

Physics
2 answers:
kifflom [539]2 years ago
4 0
Plastic would be a good insulator. Some other common insulators are glass, rubber, air, and wood.
Insulators usually protect us from the dangers of electric currents.
Hope that helped!
vovangra [49]2 years ago
3 0
Plastic is a good insulator.

Doorknob MAY be; depends on what it's made of.
You might be interested in
Calculate the average times it took the car to travel 0. 25 and 0. 50 meters. Record the averages, to two decimal places, in Tab
Illusion [34]

The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22

For the 0.50 m, the average speed will:

S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg}  = 3.13\rm \  s

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

Learn more about Average speed:

brainly.com/question/26386984

6 0
2 years ago
What is the mass of metal if it has a density of 12.459 hg/cm^3 and displaces 28.7cm^3 of water ​
natima [27]

Answer:

357.6g

Explanation:

Given parameters:

Density  = 12.459g/cm³

Volume of metal  = 28.7cm³

Unknown:

Mass of metal = ?

Solution:

The density of a substance is its mass per unit volume.

To find the mass;

     Mass of metal  = density x volume

Now insert the parameters and solve;

    Mass of metal  = 12.459 x 28.7  = 357.6g

7 0
2 years ago
Use the information below to answer questions
Ulleksa [173]

Answer:

The charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N  at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 =  2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁  = 5 × 10⁻⁸ C - q₂  (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\  = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16}  - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8}  - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C

q₁  = 5 × 10⁻⁸ C - q₂

q₁  = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁  = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

5 0
3 years ago
Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the
Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}

The force can be calculated by the formula

F=k\frac{q1q2}{(r^{\prime})^2}

Here, k is the constant whose value is

k=9\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the force will be

\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}

7 0
1 year ago
What are groups 1,2 and 3 examples of on the periodic table
pishuonlain [190]
<span>The number of the group identifies the column of the standard periodic table in which the element appears.</span>
Group 1 contains the  alkali metals ( lithium<span> (</span>Li<span>), </span>sodium<span> (</span>Na<span>), </span>potassium<span> (</span>K<span>), </span>rubidium<span> (</span>Rb<span>), </span>caesium<span> (</span>Cs<span>), and </span>francium(Fr).)<span>
Group 2 contains the alkaline earth metals (</span> beryllium<span> (</span>Be),magnesium<span> (</span>Mg<span>), </span>calcium<span> (</span>Ca<span>), </span>strontium<span> (</span>Sr<span>), </span>barium<span> (</span>Ba<span>) and </span>radium<span> (</span>Ra<span>) )
Group 3: </span><span> Scandium (Sc) and yttrium (Y) </span>
4 0
3 years ago
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