All categories

3 years ago

How does adding energy to matter by heating it affect the energy of its particles

1 answer:

6 0

When heat is added to a substance the molecules and atoms vibrate faster. As atoms vibrate faster , the space between atoms increases. The motion and spaceing of the particles determines the state of matter of the substance.

You might be interested in

. If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular

mel-nik [20]

<span>In this particular case, where car is moving through curvature, so it is moving in circular motion, force acting on car is centripetal force which holds car not to fly out. Centripetal force is always pointed in the middle of circle. Here frictional force has role of centripetal force. If frictional force is to weak, car would fly out of curvutare.</span>

7 0

2 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

3 years ago

What grade of sprain is a completely torn ligament?

Alex Ar [27]

Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.

Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.

Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.

source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/

6 0

3 years ago

If the error in the angle is 0.50, the error in sin 90o is

Sindrei [870]

At the given erro in angle, the error in the measurement of sin 90 degrees would be 0.001.

<h3>Percentage error</h3>

The percentage error of any measurement is obtained from the ratio of the error to the actual measurement.

The error of sin 90 degrees is calculated as follows;

sin 90 = 1

error in measurement = sin(90 - 0.5)

error in measurement = sin(89.5) = 0.999

<h3>Error in sin 90 degrees</h3>

Error in sin 90 degrees = 1 - 0.999

Error in sin 90 degrees = 0.001

Thus, at the given erro in angle, the error in sin 90 degrees would be 0.001.

Learn more about error in measurement here: brainly.com/question/26668346

4 0

2 years ago

Which of the following is the smallest conceivable amount of time that could pass between a lunar eclipse and a solar eclipse .

pickupchik [31]

The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
because earth is making shade on moon

in order for solar eclipse to occur the line is now:
Sun-Moon-Earth
because moon is making a shade on earth (blocking sun = solar eclipse)

Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14

5 0

3 years ago