All categories

3 years ago

How does charging by conduction occur?

2 answers:

6 0

Hello!
Charging by conduction involves the contact of a charged object to a neutral object. Suppose that a positively charged aluminum plate is touched to a neutral metal sphere. The neutral metal sphere becomes charged as the result of being contacted by the charged aluminum plate.
Hope this helped!

Nadusha1986 [10]3 years ago

6 0

Answer:

i believe it is a charged object touches a neutral object. please let me know if I'm wrong

Explanation:

You might be interested in

1+5-(-5 - 1) whats the answer

vladimir2022 [97]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re

vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = $1.8\times 10^{-5}\ \Omega\cdot m$

First, let us find the area of the circular rod.

Area is given as:

$A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2$

Now, the resistance of the material is given by the formula:

$R=\rho( \frac{L}{A})$

Express this in terms of 'L'. This gives,

$\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}$

Now, plug in the given values and solve for length 'L'. This gives,

$L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m$

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0

3 years ago

What quantity is represented by a unit called a newton (N)?

zavuch27 [327]

$\textsf {C. Force}$

$\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}$

5 0

2 years ago

I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit

Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

$M=5.97\cdot 10^{24} kg$ (mass of the Earth)

$m=3.0 kg$ (mass of the box)

$r=R=6.37\cdot 10^6 m$ (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

$F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N$

2) $g=9.8 m/s^2$

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

$\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2$

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol $g$ .

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

$F=ma$

where a is the acceleration of the object. Re-arranging,

$a=\frac{F}{m}$

which is exactly equal to the quantity we have calculated above.

5 0

3 years ago

If you were in charge of designing a wire to carry electricity across your city, state or province, which of

Anon25 [30]

Answer:

Thin, aluminium and buried underground.

Explanation:

When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.

i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.

ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.

7 0

3 years ago