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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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Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th
vitfil [10]

Answer:

q=7.65*10^{-8}C

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C

8 0
3 years ago
What is the name of the scale used to measure hurricanes
shusha [124]
Yea it’s called the Saffir-Simpson Hurricane scale, made in 1960s and further developed in 1970s
8 0
3 years ago
During a snowball fight two balls with masses of 0.4 kg and 0.6 kg, respectively, are thrown in such a manner that they meet hea
valkas [14]
Considering conservation of momentum;
m1v1 + m2v2 = m3v3

In which,
m1 = mass of snowball 1 = 0.4 kg
v1 = velocity of snowball 1 = 15 m/s
m2 = mass of snowball 2 = 0.6 kg
v2 = velocity of snow ball 2 = 15 m/s
m3 = combined mass = 1 kg
v3 = velocity after comination
Therefore;
0.4*15 + 0.6*15 = 1*v3
v3 = 6+9 = 15 m/s
KE = 1/2mv^2

Then,
KE1 = 1/2*0.4*15^2 = 45 J
KE2 = 1/2*0.6*15^2 = 67.5 J
KE3 = 1/2*1*15^2 = 112.5 J

Therefore, KE3 (kinetic energy after collision) = K1+K2 {kinetic energy before collision). And thus it is 100%.
3 0
3 years ago
Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
Lyrx [107]

The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.

<h3>Newton's Cannonball:</h3>

Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.

Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.

Learn more about Newton's Cannonball here:

brainly.com/question/18776112

#SPJ1

4 0
1 year ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
Read 2 more answers
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