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grin007 [14]
2 years ago
11

Depth of a pond seems shallower than real depth,why?​

Physics
2 answers:
IgorLugansk [536]2 years ago
7 0

Answer:

The refraction of light at the surface of water makes ponds and swimming pools appear shallower than they really are. A 1m deep pond would only appear to be 0.75 m deep when viewed from directly above. When light emerges from glass or water into air it speeds up again.

Explanation:

Viefleur [7K]2 years ago
5 0

Answer:

The property of light is the refraction of light because when light travels from a denser medium to rarer medium it bends away from the normal. The apparent depth of water is always less than the real depth of the water. Hence, water appears to be shallower than the real depth.

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How are vibration waves and energy related to sounds
Lesechka [4]

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. ... In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels.

3 0
3 years ago
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A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.
WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

 sin∅ = 3.588 × 10⁻⁷

we know that;

 sin∅  = object separation / distance from telescope

object separation =   sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷  × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0
3 years ago
A giraffe is slowly walking across a field. Is it balanced or unbalanced?
vodka [1.7K]
Your answers are correct
3 0
3 years ago
Calculate the acceleration due to gravity on the Moon. The Moon’s radius is 1.74 x 106 m and its mass is 7.35 x 1022 kg.
GarryVolchara [31]
G=?
m=7.35*10^22kg
r=1.74*10^6m
G=6.67*10^-11nm^2kg^-2
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5 0
3 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
3 years ago
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