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3 years ago

When a beam of light passes at an oblique angle into a material of lower optical density, the angle of incidence is

Physics

2 answers:

suter [353]3 years ago

7 0

Penn Foster Students: less than the angle of refraction

Anastasy [175]3 years ago

5 0

The correct answer is letter B. less than the angle of refraction. When a beam of light passes at an oblique angle into a material of lower optical density, the angle of incidence is less than the angle of refraction. The reason for that because the beam of light passes through an oblique angle.

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A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m 2, point, 0, start frac

dem82 [27]

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

7 0

2 years ago

Read 2 more answers

Write answers with significant figures: a) 17.35 g +8.498 g

Genrish500 [490]

Answer:

25.9 g

Explanation:

= 17.35

8.498

________+

= 25.848 g = 25.85 g = 25.9 g

*so sorry if wrong

6 0

2 years ago

How to find the total displacement of an object ?

OLga [1]

Answer:

it can be calculated by measuring the final distance away from a point, and then subtracting the initial distance

8 0

2 years ago

Last week we investigated the black hole at the center of our galaxy, with massMX= 8.5×1036kg. An object whose size is on the or

swat32

Answer:

$d=2.38*10^{13}m$

Explanation:

We know the mass of the hole, so we define as,

$M_x= {8.5*10^{36}}$ kg

For which centripetal force is equal to gravitational force

$\frac{mv^2}{r}=\frac{GMm}{d^2}$

Angular velocity is equal to v/r,

$w^2r=\frac{6.67*10^{-11}*8.5*10^{36}}{d^2}$

As r=1 and w=1, we clear to d,

$d^2=5.6695*10^{26}$ m

$d=2.38*10^{13}m$

6 0

3 years ago

If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m

Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is $\frac{135.9}{A}N$

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = $\frac{Pressure}{Area}$ we have

Force = $F_{120}=\frac{135.9}{A}N$ .

4 0

3 years ago