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valentinak56 [21]
3 years ago
10

An _________ is a light ray moving toward a boundary.

Physics
1 answer:
Sveta_85 [38]3 years ago
6 0

When A Light Ray Hits A Boundary,,,,, it is called refraction.

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A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
3 years ago
A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =
jarptica [38.1K]

           Kinetic energy  =  (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second.  So we'll have to
make that conversion.

        KE  =  (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

               =  (725.5) · (48 · 1000 · 1 / 3,600)²  (kg) · (km·m·hr / hr·km·sec)²

               =  (725.5) · ( 40/3 )²  ·  ( kg·m² / sec²)

               =    128,978  joules  (rounded)
8 0
3 years ago
If you exert a force of 100.0 N to lift a box a distance of 0.5 m, how much work do you do? 200 J 400 J 50 J 26 J
jeyben [28]
Work = force x distance
= 100N (force) x 0.5m (distance)
=  50J
4 0
4 years ago
Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the
Assoli18 [71]

Answer:

Part a)

\rho = 3\epsilon_0 k r^2

Part b)

Q = 4\pi \epsilon_0kR^5

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

E = kr^3 \hat r

now electric flux through a spherical surface of radius r is given as

\phi = E. A

\phi = kr^3(4\pi r^2)

now by Guass law we know that

E.A = \frac{q}{\epsilon_0}

q = 4\pi \epsilon_0kr^5

now volume charge density is given as

\rho = \frac{q}{\frac{4}{3}\pi r^3}

\rho = 3\epsilon_0 k r^2

Part b)

Total charge inside the radius R is given as

Q = 4\pi \epsilon_0kR^5

7 0
3 years ago
Air trapped in the fur of the caribou’s coat functions as _____.
ioda
It functions as insulation, to keep it warm 
5 0
3 years ago
Read 2 more answers
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