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sashaice [31]
3 years ago
13

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the poo

l from above. How deep (in cm) will it appear to be
Physics
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

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The height of a flagpole is 8.5 meters tall. If the length of the meter stick shadow is 1.5 meters long ,what would be the the l
Makovka662 [10]

Explanation:

Similar triangles:

1.5 m / 1 m = x / 8.5 m

x = 12.75 m

Round as needed.

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3 years ago
How is the sound quality of an instrument related to the overtones it produces?
Svet_ta [14]

Answer:

The fundamental tone determines what note you hear. But each instrument produces different overtones, so the blending of the fundamental tones and overtones produces different sound qualities. Resonance Resonance affects the sound quality of a. musical instrument by increasing the loudness of certain overtones.

Explanation:

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3 years ago
How to calculate force without acceleration?
Oksanka [162]
There are many ways to calculate force. However, they are completely different aspects. The force on an area is pressure, so F = p*A. In magnetic fields you use another formule (forgot the formule atm). Unless you define what kind of force you are looking for, I can't help you any further.
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2 years ago
Read 2 more answers
A street light is on top of a 12 foot pole. a person who is 5 feet tall walks away from the pole at a rate of 4 feet per second.
Karo-lina-s [1.5K]
<span>2 and 6/7 ft/sec or approximately 2.857 ft/sec First, let's make an expression to express the length of the person's shadow as a function of the distance from the light pole. Variables d = distance person is from pole s = length of shadow person is casting. The right triangle that the person makes with their shadow is similar to the right triangle made by the light pole and the end of the person's shadow. So 5/12 = s/(s+d) Now solve for s in terms of d 5/12 = s/(s+d) 5/12(s+d) = s 5s/12 + 5d/12 = s 5d/12 = s - 5s/12 5d = 12s - 5s 5d = 7s 5d/7 = s So for every foot the person walks away from the pole, the shadow increases by 5/7ths of a foot. Since the person is moving at 4 ft/sec, just multiply 5/7 by 4. 5/7 * 4 = 20/7 = 2 6/7 or approximately 2.857142857 ft/sec</span>
3 0
2 years ago
A rock drops from the top of a 10.5 m building. What is the velocity when he hits the ground?what is the Plot the position, velo
Crank

Answer:

When the rock is on top of the building, it does not move, so it only has potential energy.

The potential energy can be written as:

U = m*g*h

where m is the mass, g is the gravitational acceleration, h is the height.

Now, as the rock starts to fall down, the potential energy transforms into kinetic energy.

K = (m/2)*v^2

Where v is the velocity.

When the rock hits the ground, all the potential energy has ben converted into kinetic energy, then:

U = K

m*g*h = (m/2)*v^2

Here we can isolate v:

v = √(2*g*h)

and g = 9.8m/s^2

      h = 10.5m

v = √(2*10.5m*9.8m/s^2) = 14.34m/s

Now the second question:

"what is the Plot the position, velocity and acceleration vs. time"

I suppose that you need to select the correct plot for each thing, the images are not given, so let's analyze how each plot is:

The motion equations are:

Acceleration:

Here we have only the gravitational acceleration, so we can write:

a(t) = -g

This is a constant, the graph will be a horizontal line at y = -g.

Velocity:

We integrate the acceleration over time, the constant of integration is the initial velocity, that in this case is zero.

v(t) = -g*t

This is a linear equation with slope equal to -g, and y-intercept equal to zero.

Position.

We integrate again over time, this time the constant of integration will be the initial height of the rock = 10.5m

The equation is:

p(t) = -(g/2)*t^2 + 10.5m

This is a quadratic equation with a negative leading coefficient, so the arms go downwards.

4 0
3 years ago
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