1) A star burst can be defined as stars formed from recycled dead star materials.

Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th

puteri [66]

Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.

6 0

2 years ago

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Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This r

alexandr1967 [171]

Answer:

to create the particle the speed must be greater than 2.25 10⁸ m / s

Explanation:

In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium

n = c / v

where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio

in this case they give us that the medium matter water them that has a refractive index of

n = 1,333

we clear

v = c / n

let's calculate

v = 3 10⁸ / 1,333

v = 2.25 10⁸ m / s

to create the particle the speed must be greater than 2.25 10⁸ m / s

6 0

2 years ago

The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on

choli [55]

Answer:

25 times the average speed

7 1

2 years ago

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A hunter wishes to cross a river that is 1.5 km wide and flows with a velocity of 5.0 km/h parallel to its banks. The hunter use

anzhelika [568]

Answer:137.48 s

Explanation:

Given

Width of river=1.5 km

velocity of river=5 km/h

velocity of boat w.r.t river =12 km/h

To cross the river in minimum time hunter needs to cross the river perpendicular to the flow

i.e. velocity of boat w.r.t water must be perpendicular

i.e. x component of boat must be equal to river flow

$12cos\theta =5$

where $\theta$ is angle made by boat w.r.t bank

$cos\theta =\frac{5}{12}$

$cos\theta =0.416$

$\theta =65.417^{\circ}$

its vertical component is $12sin(65.417)=10.91 m/s$

Time taken $=\frac{1.5\times 1000}{10.91}=137.48 s$

6 0

2 years ago

In a race, a runner traveled 12 meters in 4.0 seconds as she accelerated uniformly from rest. The magnitude of the acceleration

mestny [16]

The runner is moving by uniformly accelerated motion, starting from rest (so, his initial velocity is zero). The law of motion of the runner is
 $x(t) = \frac{1}{2} at^2$ 
where x(t) is the distance covered after time t, and a is the acceleration of the runner. By re-arranging the formula, we get
 $a= \frac{2S}{t^2}$ 
We know the runner has covered a distance of S=12m in t=4.0 s, and if we plug these numbers into the equation, we find the acceleration of the runner:
 $a= \frac{2S}{t^2} = \frac{(2*12m)}{(4s)^2} =1.5 m/s^2$

5 0

2 years ago