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faust18 [17]
2 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]2 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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Answer:

stratosphere

Explanation:

the stratosphere is the second major layer of the atmosphere just above the troposphere, it has its lower area to be cooler and higher area to be warmer due to the ozone layers absorption of ultraviolet rays. It is also the layers which planes fly through because it is stable.

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3 years ago
A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k
Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

I_1=m_1r^2

I_1=7\times (0.25)^2=0.437\ kgm^2

The moment of inertia of the rod about one end is given by :

I_2=\dfrac{m_2l^2}{3}

l = r

I_2=\dfrac{m_2r^2}{3}

I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2

For 6 spokes, I_2=0.025\times 6=0.15\ kgm^2

So, the net moment of inertia of the wagon is :

I=I_1+I_2

I=0.437+0.15=0.587\ kgm^2

So, the moment of inertia of the wagon wheel for rotation about its axis is 0.587\ kgm^2. Hence, this is the required solution.

4 0
3 years ago
A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

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The weight is 45 N, because the three chains hold the sign, and each contributes 15 N.

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