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3 years ago

Waves travel as "packets" of several waves. in these "packets," a wave travels at __________.

Physics

2 answers:

dalvyx [7]3 years ago

8 0

Answer:

a. the same speed as the wave energy

Explanation:

Wave packet is only an envelope of a more localized wave. The entire wave is a combination of the discrete wave packets.

Wave is any disturbance that travels across space. A wave does not carry the particles of the medium in which travels. It carries energy from one point to another as it travels.

For an electromagnetic wave, the energy of the wave is due to the vibrations of the electric and magnetic fields of the wave. For sound waves, wave energy is the vibrations or disturbance of the particles of the medium in which the sound wave travels.

madam [21]3 years ago

4 0

A. the same speed as the wave energy

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A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes

Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0

3 years ago

Starting from rest, approximately how far does an object fall freely in 3 seconds?

Sergio039 [100]

Hi! In 3 seconds the object will fall approximately 44 meters.

8 0

3 years ago

What storm causes flash floods? <br> Tornado <br> Thunderstorm<br> Hurricane

Mariulka [41]

Answer:

hurricane.

Explanation:

............................

8 0

3 years ago

is dimensionally correct relation necessarily to be a correct physical relation? explain with example.

Andreas93 [3]

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

4 0

3 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

kozerog [31]

Answer:

v’= 279.66 m / s

Explanation:

We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.

Initial instant. Before the crash

p₀ = m v₀ + 0

Final moment. After the crash

p_f = m v + M v ’

how the tidal wave is preserved

p₀ = p_f

m v₀ = m v + M v ’

v = $\frac{m v_o - Mv'}{m}$

let's calculate

v ’= $\frac{0.00467 \ 619 - 0.072 \ 22}{0.004676}$

v ’= $\frac{2.89- 1.584}{ 0.00467}$

v ’= 279.66 m / s

4 0

3 years ago