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3 years ago

Which is the correct representation of the right-hand rule for a current flowing to the right?

Physics

2 answers:

Elza [17]3 years ago

7 0

Answer:

The third image

Explanation:

The one with the thumb pointing to the right

SVETLANKA909090 [29]3 years ago

4 0

Answer:

3, correct on Edge 2020

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When atoms of an element are excited, they emit specific wavelengths of light. How is this similar to a fingerprint when Fraunho

Anika [276]

Answer:

As you may know, each element has a "fixed" number of protons and electrons.

These electrons live in elliptical orbits around the nucleus, called valence levels or energy levels.

We know that as further away are the orbits from the nucleus, the more energy has the electrons in it. (And those energies are fixed)

Now, when an electron jumps from a level to another, there is also a jump in energy, and that jump depends only on the levels, then the jump in energy is fixed.

Particularly, when an electron jumps from a more energetic level to a less energetic one, that change in energy must be compensated in some way, and that way is by radiating a photon whose energy is exactly the same as the energy of the jump.

And the energy of a photon is related to the wavelength of the photon, then we can conclude that for a given element, the possible jumps of energy levels are known, meaning that the possible "jumps in energy" are known, which means that the wavelengths of the radiated photons also are known. Then by looking at the colors of the bands (whose depend on the wavelength of the radiated photons) we can know almost exactly what elements are radiating them.

7 0

3 years ago

A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t

Lapatulllka [165]

I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.

Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)

-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.

-- So at the top of its trajectory, its KE is 0.25 of what it had originally.

That's E/4 .

3 0

2 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

2 years ago

What happens if sulfur forms an ionic bond with another element

bija089 [108]

Answer:

The sulfur will accept the electrons

Explanation:

8 0

3 years ago

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

Trava [24]

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

$T=2\pi \sqrt{\frac{m}{k} }$

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

6 0

7 months ago