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faust18 [17]
3 years ago
5

A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm

under the surface of the water.
A) What is the apparent depth of the fish when viewed at normal incidence?
B) What is the apparent depth of the image of the fish when viewed at normal incidence?
Physics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish  = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.

(B). the depth of the fish in the mirror.

d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm

Now find the depth of reflection of the fish in the bottom of the tank.

d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8  = 24.66\\

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lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is  27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus 7=\frac{x}{v}----1

for return trip

4=\frac{x}{v+27}-----2

divide  1 & 2

\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}

7v=4v+4\cdot 27

3v=4\cdot 27

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3 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




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• using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use.

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