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NARA [144]
2 years ago
8

Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish

really is? Does the fish see the person's face closer or farther than it really is? Explain your answer.
Physics
1 answer:
Sunny_sXe [5.5K]2 years ago
4 0

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

        θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

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Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th
katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0
1 year ago
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Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

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2 years ago
Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
A compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible fri
enot [183]

(a) 80 N/m

The spring constant can be found by using Hooke's law:

F=kx

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m

(b) 0.84 Hz

The frequency of oscillation of the system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

v=\omega A

where

\omega is the angular frequency

A is the amplitude of the motion

For this system, we have

\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s

A=0.200 m (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

v=(5.25 rad/s)(0.200 m)=1.05 m/s

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

a=\omega^2 A

Using the numbers we calculated in part c):

\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s

A=0.200 m

we find immediately the maximum acceleration:

a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2

(f) At the position of maximum displacement: x=\pm 0.200 m

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

a=\frac{F}{m}

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

F=kx

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

E=K=\frac{1}{2}mv_{max}^2

where

m = 2.90 kg is the mass

v_{max}=1.05 m/s is the maximum speed

Solving for E, we find

E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

x=\frac{1}{3}(0.200 m)=0.0667 m

so the elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J

and since the total energy E = 1.60 J is conserved, the kinetic energy is

K=E-U=1.60 J-0.18 J=1.42 J

And from the relationship between kinetic energy and speed, we can find the speed of the system:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

x=\frac{1}{3}(0.200 m)=0.0667 m

So the restoring force exerted by the spring on the mass is

F=kx=(80 N/m)(0.0667 m)=5.34 N

And so, we can calculate the acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2

8 0
1 year ago
100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:
love history [14]

Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

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