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trasher [3.6K]
3 years ago
8

Materials that allow electricity to pass through them are called ?

Physics
1 answer:
lianna [129]3 years ago
7 0

Answer:

Materials that allow electricity to pass through them are called conductors.

Copper wire is a good conductor. Materials that do not allow electricity to pass through them are called insulators.

Explanation:

Can I have brainliest?

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The law of conservation of matter states that during a chemical reaction, the amount of matter ​
mixas84 [53]

Answer:

The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.

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3 years ago
What is the mass of a 5000 N car on Earth?
puteri [66]

Answer:

510.2kg

Explanation:

use w=mg then put the value you will get answer

4 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of it
White raven [17]

Answer:

The positions are  0.0194 m  and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

v = \omega \sqrt{A^2-x^2}

the maximum speed of the simple harmonic motion is given as;

v_{max} = \omega A

when the speed equal one fourth of its maximum speed

v =\frac{v_{max}}{4}

\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m  \ \ or\  - 0.0194  \ m

Thus, the positions are  0.0194 m and - 0.0194 m.

8 0
3 years ago
Which of the following is the brightness of a star as we see if from Earth?
My name is Ann [436]
I think that it is apparent magnitude
8 0
3 years ago
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

7 0
4 years ago
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