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Verizon [17]
3 years ago
13

According to the principal of wave refraction, when light from the atmosphere enters a medium of greater density at an oblique a

ngle its _____________. a. velocity speeds up but angle remains the same b. velocity remains the same but angle changes c. velocity slows down but angle remains the same d. velocity slows down and angle changes
Physics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

d. velocity slows down and angle changes

Explanation:

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I HAVE A PHYSICS TEST, ITS 25 QUESTIONS AND I HAVE ABOUT AN HOUR TO SOLVE IT. PLEASE IF YOU’RE GOOD AT PHYSICS CONTACT ME ASAP
Katena32 [7]

Answer:

in which Class u studying

5 0
3 years ago
Read 2 more answers
Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field
Scilla [17]

Answer:

Electric field lines point away from positive charges and toward negative charges. <u>True</u>

Electric field lines are continuous; they do not have a beginning or an ending.<u> False</u>

Electric field lines can never intersect. <u>True</u>

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

Explanation:

Electric field lines point away from positive charges and toward negative charges. Always the field lines go to negative charges and leave from positive charges.

Electric field lines are continuous; they do not have a beginning or an ending.<u> False  </u>

Because the field lines starts at positive charges and ends in negative charges.

Electric field lines can never intersect. <u>True</u>

It can not intercept the field lines because in that point the the field would have two directions. Besides, in that point the real value of the field should be found adding both field lines.

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

This fact is opposite to that so the regions of space where the magnitude the electric field is weak the lines are father apart and where the field is strong  the lines are close together.

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

This statement correspond to the definition of the field line.

5 0
3 years ago
Starting at t = 0 a net external force in the +x-direction is applied to an object that has mass 5.00 kg. A graph of the force a
Reptile [31]

Answer:

  15√2 N

Explanation:

The acceleration is given by ...

  a = F/m = 5t/5 = t . . . . meters/second^2

The velocity is the integral of acceleration:

  v = ∫a·dt = (1/2)t^2

This will be 9 m/s when ...

  9 = (1/2)t^2

  t = √18 . . . . seconds

And the force at that time is ...

  F = 5(√18) = 15√2 . . . . newtons

6 0
3 years ago
PLEASE ANYONE CAN HELP ME !E.x/A block of metal has a volume of 0.09 m3
LuckyWell [14K]

Answer:

B = 1058.4  N

Explanation:

Given that,

The volume of a metal block, V = 0.09 m³

The density of fluid, d = 1200 kg/m³

We need to find the buoyant force when it's Completely  immersed in brine. The formula for the buoyant force is given by :

B=\rho gV

g is acceleration due to gravity

B=1200\times 9.8\times 0.09\\\\B=1058.4\ N

So, the required buoyant force is 1058.4  N.

3 0
3 years ago
A competitive go-cart driver is traveling 32 m/s. He sees a caution flag go up, so he slows at a rate of -1.5 m/s^2 in 10.8 s. W
Svetllana [295]

Answer:

15.8 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = [final velocity (v) – initial velocity (u)] / time (t)

a = (v – u) /t

With the above formula, we can obtain the final velocity of go-cart driver as follow:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

a = (v – u) /t

– 1.5 = (v – 32) / 10.8

Cross multiply

(v – 32) = –1.5 × 10.8

v – 32 = – 16.2

Collect like terms

v = – 16.2 + 32

v = 15.8 m/s

Therefore, the final velocity of go-cart driver is 15.8 m/s.

5 0
3 years ago
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