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3 years ago

For a transverse wave, what is a wavefront?

Physics

2 answers:

Natalka [10]3 years ago

8 0

Answer:

A wavefront is the long edge that moves, for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v.

spin [16.1K]3 years ago

5 0

Answer:

A wavefront is a surface containing points affected in the same way by a wave at a given time such as crest and troughs

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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth

mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

4 0

3 years ago

Read 2 more answers

A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free

Maslowich

Consider that the bar magnet has a magnetic field that is acting around it, which will imply that there is a change in the magnetic flux through the loop whenever it moves towards the conducting loop. This could be described as an induction of the electromotive Force in the circuit from Faraday's law.

In turn by Lenz's law, said electromotive force opposes the change in the magnetic flux of the circuit. Therefore, there is a force that opposes the movement of the bar magnet through the conductor loop. Therefore, the bar magnet does not suffer free fall motion.

The bar magnet does not move as a freely falling object.

3 0

3 years ago

DEFINE REFRACTION? describe how a ray of light is refracted when it passes through a glass block

jek_recluse [69]

When a ray passes from air into glass the direction in which the light ray is travelling changes. The light ray appears to bend as it as it passes through the surface of the glass. ... This 'bending of a ray of light' when it passes from one substance into another substance is called refraction.

8 0

3 years ago

A bicyclist is travelling at 25 m/s when he begins to decelerate at -4m/s2 . How fast is travelling after 5 seconds

Umnica [9.8K]

Initial velocity (Vi) = 25 m/s

acceleration (a) = $-4 m/s^{2}$

time interval (t) = 5 sec

let us assume that final velocity after 5 sec be Vf

As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. $V_{f} = V_{i} + at$

Hence, $V_{f} = 25 +(-4)(5) = 25 -20 = 5 m/s$

so, the velocity of bicyclist will be 5 m/s after 5 sec

7 0

3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$