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ira [324]
3 years ago
10

Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

ll perceive the liquid to be what apparent depth?
Physics
1 answer:
Tju [1.3M]3 years ago
7 0

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

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In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g
Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η=\frac{electric power}{water power}

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0
2 years ago
An imaginary line perpendicular to a reflecting surface is called _________.
n200080 [17]
<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>
5 0
2 years ago
Read 2 more answers
Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet
telo118 [61]

Answer:

- Bigger mass of planet B  

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

v=\sqrt{\frac{GM}{r}}

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B  

- orbiting closer to planet B

6 0
3 years ago
Each tyre of a car has an area of 100 cm2 in contact with the ground. The car has a mass of 1600 kg. The weight of the car is eq
vlabodo [156]

Answer:

b is the answer

Explanation:

tq friend b is the answer

3 0
2 years ago
Read 2 more answers
Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The
AnnyKZ [126]

Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y:  opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N  directed down

for equilibrium system

Fy= Ty=107.2 N

Fy=107.2 N upward component of the force acting on the stake

8 0
2 years ago
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