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3 years ago

Which of the following distinguishes electromagnetic waves from mechanical waves

Physics

1 answer:

Ivan3 years ago

6 0

Answer:

Explanation:

1. Mechanical waves require material medium for their propagation while electromagnetic waves do not require material medium for their propagation.

2. Mechanical waves can either be transverse or longitudinal while electromagnetic waves are transverse.(Transverse waves are waves in which the vibration of the particules of the medium is perpendicular to the direction of the motion of wave. E.g water waves, waves of a plucked string and all electromagnetic waves RIVUXG . Longitudinal waves are waves whose vibration are parallel to the direction of the motion of the medium e.g waves in strings, sound waves.e.t.c)

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Answer:

i think it would be Galvanometer

Explanation:

because it would have to be a type of coil or wire

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16. Cassandra notices that when she breathes on a cool window, the water vapor in her breath forms

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2 years ago

Which of the following is the best reason to rest muscle groups between workouts?

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Answer;

It allows the muscles time to heal.

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Lake point tower in chicago is the tallest apartment building in the united states. suppose you take the elevator from street le

notka56 [123]

Using the idea of work done under gravity, the height of the building is 187 m.

<h3>Work done in a gravitational field</h3>

We must recall that the work done in a gravitational field is given by; mgh

m= mass

g = acceleration due to gravity

h = height

mass = 60.0 kg

Workdone = 1.15x10^5 J

W = mgh

h = W/mg

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Learn more about work done: brainly.com/question/13662169?

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2 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

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3 years ago