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2 years ago

What type of wave cannot travel in a vacuum

Physics

2 answers:

AlexFokin [52]2 years ago

7 0

The Answer is Sound Waves because in the vacuum of space, there is no medium to transmit of these mechanical waves.

lions [1.4K]2 years ago

5 0

A sound wave. Because in a vacuum there is no medium in a vacuum. And the only wave that requires a medium to travel through is a sound wave.

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For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

d = 8 furlongs

It traveled 8 furlongs for the first 4.0 seconds.

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2 years ago

A lawn sprinkler sprays water 8 feet at full pressure as it rotates 360 degrees. if the water pressure is reduced by 50%, what i

lina2011 [118]

The area of the sprinkles can be determined through the area of a circle that is pi * r^2 in which the given dimensions above are the radii, r. The second scenarios radius is only half of the original, that is 4 ft. In this case, we can compute the area of the second again. We calculate next the difference of two areas of circles.

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2 years ago

Given the indices of refraction n1 and n2 of material 1 and material 2, respectively, rank these scenarios on the basis of the p

lisov135 [29]

Answer:

$c>d>f=a>b>e$

Explanation:

When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.

According to the Snell's law:

$\rm refractive\ index\ (n)=\frac{speed\ of\ light\ in\ the\ incident\ medium}{speed\ of\ light\ in\ the\ refracted\ medium}$

$n_1-n_2=1.33-1.00\\=0.33$

$n_2-n_1=1.46-1.33$

$=0.23$

$n_2-n_1=2.42-1.33\\=1.09$

$n_2-n_1=1.46-1.00\\=0.46$

$n_1-n_2=1.50-1.33\\=0.17$

$n_2-n_1=1.33-1.00\\=0.33$

$c>d>f=a>b>e$

5 0

2 years ago

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

$F=\dfrac{mv^2}{r}$