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3 years ago

What are some examples of EM waves

Physics

1 answer:

Alexxandr [17]3 years ago

5 0

Answer:

Microwaves
Heatwaves
Ultraviolent waves
WiFi signal
4G/5G signals

Explanation:

All of these are examples of electromagnetic radiation or energy.

The spectra comprises most energy and matter we can account for so far.

The main distinction between EM waves and for example longitudinal acoustic waves is the medium of energy transfer (and rules). EMWs are subject to the idiosyncratic effects of Quantum Mechanics while other 'macro' waves whose medium relies on gross material, are apparently not.

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A force of 5.00 N to the left causes a 1.35 kg book to have a net acceleration of 0.76 m/s2 to the left. What is the frictional

Nuetrik [128]

Answer:

4.0 N

Explanation:

Sum the forces in the x direction:

∑F = ma

F − Fr = ma

Fr = F − ma

Fr = 5.00 N − (1.35 kg) (0.76 m/s²)

Fr = 4.0 N

8 0

3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$