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hammer [34]
3 years ago
6

Where would you expect light waves to move faster?

Physics
2 answers:
KengaRu [80]3 years ago
7 0

Answer:

a

Explanation:

vlabodo [156]3 years ago
6 0
I say in space which is number A. That is because there is nothing that would restrict anything in space and it is quite open and free. Hope this helped.
You might be interested in
If scientists could create a sheet of glass with a refractive index so high that it took light months or years to travel through
omeli [17]

Wow ! They could set up sheets of "slow glass" beside beautiful forests with rivers and squirrels and deer and grassy fields, and load a year of this scene into the glass, and then sell it to people who live next to dirty brick walls or ugly empty lots, and those people could install the slow glass in their windows and have beautiful scenery, until it all worked its way out of the glass.

This is a great idea ! If you possibly can, find and read the sci-fi short story "Light of Other Days" written by Bob Shaw, published in Analog Science Fiction in 1966. It's all about this exact type of glass. I read this story in 1966 and I never forgot it ! (Not yet anyway.)

4 0
3 years ago
A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend
GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

a=\frac{h}{5(t_1)^2}

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = y = \frac{1}{2}a(t_1)^2

Velocity = v_1=at_1

<h3>Stage 2</h3>

Constant velocity where

Velocity = v_o=v_1=at_1

Distance =

<h3>y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\</h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = v_0=v_1=at_1

Distance =

y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = \frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

a=\frac{h}{5(t_1)^2}

6 0
3 years ago
Basketball is my favorite sport!
WARRIOR [948]
Center.........................
6 0
3 years ago
Read 2 more answers
A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum
AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, \omega=33\frac{1}{3} rev/min , radius of the record r=0.1m and the distance between any two successive bumps on the groove as d=1.75mm.

The linear speed of the record in meters per second is:

v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\

#From v above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

Hits/second=v/d    \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199

Hence the bumps hit the stylus at around 199hit/s

8 0
3 years ago
What is the magnitude of the torque that the axle must apply to prevent the disk from rotating?
mihalych1998 [28]

The required torque at the axle, is given by the difference between the

moments of the applied forces.

The torque required is <u>19.62 N·m counterclockwise</u>

Reasons:

The given parameters are;

Mass of the disk, m = 5.0 kg

Location of the axle = Half the radius of the disk

Diameter of the disk, D = 40 cm = 0.4 m

Applied mass, 0.1 m from the axle = 15 kg

Applied mass, 0.3 m from the axle = 10 kg

Required:

Magnitude of torque at the axle that prevent the disk from rotating

Solution:

Torque needed = Clockwise moment - Counterclockwise moment

Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m

Counterclockwise moment = 15 kg × 0.1 m  × 9.81 m/s² = 14.715 N·m

τ + Counterclockwise moment = Clockwise moment

τ + 14.715 N·m = 34.335 N·m

Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m

Torque required, τ = <u>19.62 N·m counterclockwise</u>

Learn more here:

brainly.com/question/19044661

brainly.com/question/19247046

<em>The probable question drawing obtained from a similar question online is attached</em>

7 0
2 years ago
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