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3 years ago

What's the difference between electromagnetic waves and electromagnetic spectrum?

1 answer:

4 0

The difference between the two is, well for one

Spectrum: The entire range that the "<em>waves" </em>could be such, as visible light, x-ray's and so on.

Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.

<em>It may confuse you but it makes sense to me (Sorry)</em>

You might be interested in

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

The mass is $m_2 =21.75*10^{-27} \ kg$

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago

A rock falls off a 15 meter cliff. How fast is the rock traveling vertically two seconds later?

oksian1 [2.3K]

The acceleration of gravity on Earth is 9.8 m/s². That means that
an object falling under the influence of gravity will move 9.8 m/s
faster than it was moving a second earlier.

Falling from rest, it will be moving 9.8 m/s after the first second,
and 19.6 m/s after the 2nd second.

The height from which it fell doesn't matter.

8 0

3 years ago

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction

nikitadnepr [17]

Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m