All categories

2 years ago

What is the relationship between electricity and magnetism? Choose the best answer.

Physics

1 answer:

Anastaziya [24]2 years ago

5 0

Answer:

Explanation:

Magnetism is form of electricity.

You might be interested in

Chang wants to build a circuit that will light up the bulb from a flashlight. Which component could Chang leave out and still li

Elan Coil [88]

The component that could Chang leave out and still light up the bulb is the switch.

4 0

2 years ago

Read 2 more answers

A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh

lianna [129]

Answer: 3.21 N

$Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}$

$\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg$

For weight, we will multiply by $g=9.8 m/s^{-2}$

$weight= 0.328\times9.8=3.21\hspace{1mm}N$

Hence, the rock would weigh 3.21 N.

3 0

2 years ago

Atmospheric pressure and density at the top of Mount Everest

Art [367]

The summit of Mount Everest has an average pressure around 30 kPa. ... A barometer also measures variations in atmospheric pressure. As altitude increases, the air becomes thinner, the density of air decreases, and the pressure of the air decreases as well.

3 0

3 years ago

The graph represents velocity over time...

Nady [450]

Answer:

where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please

8 0

2 years ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

6 0

2 years ago