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irakobra [83]
3 years ago
8

A piece of magnetic material that retains its magnetic properties after it is removed from a magnetic field.

Physics
2 answers:
NemiM [27]3 years ago
7 0

Answer: d

Explanation: Unscrew one light, if the others remain on it is a parallel circuit.

Anna007 [38]3 years ago
5 0

Answer:

Ferromagnetic materials

Explanation:

Ferromagnetic materials are highly susceptible to an external magnetic field. They show a strong magnetic field attraction and can maintain their magnetic properties while suppressing the external field.

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What are the SI units of thermal conductivity?​
kotykmax [81]

Answer:

The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).

Explanation:

hope this will help u

7 0
3 years ago
Water is an example of ___________, while hydrogen is an example of __________.
-BARSIC- [3]
Are you referring to the fact that water is a compound while hydrogen is an element? If I'm wrong just comment and clarify and I can edit it, I don't even know what kind of unit you're in. :)
3 0
3 years ago
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Which one of the following accurately describes the force of gravity?
elena55 [62]
Choice ' C ' is a true statement.
The other choices aren't.
8 0
3 years ago
You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag
elena-14-01-66 [18.8K]

Answer:

Velocity=6.03m/s

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

So

A^{2}+B^{2}=C^{2}\\  C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m

Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s

7 0
3 years ago
In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
3 years ago
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