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spayn [35]
3 years ago
5

A coin is resting on the bottom of an empty container. The container is then fi lled to the brim three times, each time with a d

iff erent liquid. An observer (in air) is directly above the coin and looks down at it. With liquid A in the container, the apparent depth of the coin is 7 cm, with liquid B it is 6 cm, and with liquid C it is 5 cm. Rank the indices of refraction of the liquids in descending order (largest fist).
a. nC, nA, nB
b. nA, nB, nC
c. nB, nA, nC
d. nC, nB, nA
e. nA, nC, nB
Physics
1 answer:
ahrayia [7]3 years ago
3 0

when we placed in the container of real depth assuming = d

if the container filled with liquid A then its apparent depth d' = 7cm

so the refractive index nA = real depth / apparent depth

                                       = d/7cm = 0.1428d

if the container filled with liquid B then its apparent depth d' = 6cm

so the refractive index nB = real depth / apparent depth

                                       = d/6cm = 0.166d

if the container filled with liquid C then its apparent depth d' = 5cm

so the refractive index nC = real depth / apparent depth

                                       = d/5cm = 0.2d

Since the refractive index is inversely proportional to the apparent depth

then the refractive indices are nC > nB > nA

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A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

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You would see the darkest part of the moon
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Pls help i’ll give brainliest if you give a correct answer!!
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Answer:

C. The distance traveled by an object at a certain velocity.

Explanation:

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Decay of ______ waste present in waste water leads to unpleasant smell
victus00 [196]

Decay of Schizomycetes waste present in waste water leads to unpleasant smell.

These microorganisms, known as iron and sulfur bacteria, cause hydrogen sulfide to be found in water supplies. They include crenothrix and beggiatoa. Most often found in groundwater supplies, they produce an offensive odor of decaying matter. (I'm not sure)

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The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO.
Genrish500 [490]

Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

Emf=-NdΦ/dt

Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

i=-0.96NCos(120πt) Ampere

d. Power delivered to the loop

Power is given as

P=IV

P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm

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3 years ago
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