1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GuDViN [60]
3 years ago
11

An underwater diver sees the sun 50° above horizontal. how high is the sun above the horizon to a fisherman in a boat above the

diver?
Physics
1 answer:
Usimov [2.4K]3 years ago
3 0
Snell's law states that:

n1 Sin∅1 = n2 Sin ∅2

Where, medium 1 with (n1 = 1.33) is water and medium 2 with (n2 = 1) is the air, ∅1 = 90-50 = 40°

Therefore,
Sin ∅2 = n1/n2 *Sin ∅1 = 1.33/1 *Sin 40 = 0.4833=> ∅1 = Sin ^- (0.4833) = 28.9 °

The fisherman the sun at 61.1° (90-∅2) above the horizontal.
You might be interested in
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
A cube that has a volume of 1.00 m³ contains N distinguishable particles.
grandymaker [24]

Answer:

1 P = 0.5

2 P = 0.3

3 P = 0.01

Explanation:

The probability formula is

          P =V^N

Where P is the probability  V is the volume while N is the number of distinguishing particles

So for N = 1 and V = 0.500m^3

           P = (0.500m^3)^1\\

               = 0.5

For  N = 1 and V =  0.300m^3

           P = (0.300m^3)^1

               = 0.3

For   N = 1 and V = 0.0100m^3

     P =(0.0100m^3)^1

         = 0.01

         

6 0
3 years ago
Which term describes an image that is in the opposite orientation than the object from which it was formed? virtual erect invert
DiKsa [7]

Answer:

lateral

...............

7 0
3 years ago
What happens during the process of deposition
frez [133]
Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .
8 0
3 years ago
The planet Saturn has a mass that is 95 times Earth's mass and a radius that is 9.4 times Earth's radius. What is the accelerati
ziro4ka [17]

Answer:

10.55111 m/s²

Explanation:

M = Mass of Saturn = 95\times 5.972\times 10^{24}\ kg

r = Radius of Saturn = 9.4\times 6.371\times 10^6\ m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 95\times 5.972\times 10^{24}}{(9.4\times 6.371\times 10^6)^2}\\\Rightarrow g=10.55111\ m/s^2

The acceleration due to gravity on Saturn is 10.55111 m/s²

4 0
3 years ago
Other questions:
  • Which type of central heating system involves fans and ducts to circulate warm air?
    11·2 answers
  • If gravity on the moon is 1.62m/s, what is the weight of a 200g rock on the moon
    10·1 answer
  • A tennis player hits a 1.45 kg tennis ball with a racket of mass 2.5 kg. If he hits the ball with a velocity of 7.5 m/s and then
    11·1 answer
  • Describe the work performed by a ski lift in terms of kinetic and gravitational potential energy
    9·2 answers
  • Which statement best describes why the Milky Way is shaped a certain way?
    5·1 answer
  • Why is physics worth knowing about
    10·1 answer
  • Which of the following is a unit of acceleration?
    11·1 answer
  • The mass of Earth is 5.972×1024 kg and its orbital radius is an average of 1.496×1011 m . Calculate its linear momentum.
    6·1 answer
  • 5. Use of geothermal energy is not is not possible to generate electricity in Nepal​
    9·1 answer
  • How much time does it take the cheetah to travel 500 meters, if its average speed is 70 meters per second?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!