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Triss [41]
4 years ago
7

A submarine is 3.00 x 10^2 m horizontally from shore and 120.0 m beneath the surface of the water. A laser beam is sent from the

submarine so that the beam strikes 2.10 x 10^2 m from the buildingstanding on the shoreand the laser beam hits the target on the top of the building. Use n=1.333 for the refractive index of water.A)Calculate the angle of incidence ?1 on the water/air interface.
B)Calculate the angle of refraction ?2.
C)Find angle ? with the horizontal.
D)Find the height h of the building
Physics
1 answer:
PSYCHO15rus [73]4 years ago
3 0

Answer:

Explanation:

a ) Let angle of incidence and angle of refraction be i and r respectively .

submarine is 300 m away from the shore . The point where laser strikes the surface of sea is 90 m horizontally away .

Tan r = 90 / 120

= 3 / 4

.75

r = 37 degree

c ) sini / sin37 = 1.333

sini = .8

i = 53 degree

Tan 53 = 210 / h , h is height of the building .

h = 210 / tan 53

= 158 m

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A ball dropped onto a trampoline returns to the same height after the rebound.
Jet001 [13]

a)The ball's potential energy changes as it moves from the release point to the top of the rebound.

b)The ball's potential energy is at its highest and rests when it is in the top position.

<h3>What is potential energy?</h3>

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

The potential energy is mainly dependent upon the height of the object.

Potential energy = mgh

The kinetic energy of the body is due to the virtue of motion.

According to the Law of Conservation of Energy, energy can neither be created nor destroyed but can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

A ball dropped onto a trampoline returns to the same height after the rebound.

Hence, the ball's potential energy changes as it moves from the release point to the top of the rebound. When the ball is at the top position the potential energy is maximum and the ball is at rest.

To learn more about the potential energy, refer to the link;

brainly.com/question/24284560

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7 0
2 years ago
Particles of m1 and m2 (m2&gt;m1) are connected by a line in extensible string passing over a smooth fixed pulley. Initially, bo
Tresset [83]

Answer:

The velocity with which the mass will hit the floor is v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.}

Explanation:

If the tension in the string is T, for m_1 we have

T- m_1g =m_1a,

and for the mass m_2

T -m_2g = -m_2a

From these equations we solve for a and get:

a =(\dfrac{m_2-m_1}{m_2+m_1}) g.

The kinematic equation

v_f^2 = v_0^2+2ax

gives the final velocity v_f of a particle, when its initial velocity was v_0, and has traveled a distance x while undergoing acceleration a.

In our case

v_0 = 0 (the initial velocity of the particles is zero)

a =(\dfrac{m_2-m_1}{m_2+m_1}) g.

which gives us

v_f^2 = 2ax

v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g

\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }

which is the velocity with which the mass m_2 will hit the floor.

8 0
3 years ago
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4 years ago
A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
Zinaida [17]

Answer:

y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}

Explanation:

Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles,  The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

work done by the spring force = ¹/₂k(y₀² - y²)

work done by friction = - f(y - y₀)

work done by gravity = mg(y - y₀)

kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)

So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0

¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

¹/₂ky² + fy - mgy = 0

Using the quadratic formula

y = \frac{-f +/- \sqrt{f^{2} - 4 X\frac{k}{2} X -mg}}{2 X \frac{k}{2} }\\ y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}

4 0
3 years ago
A circular copper loop is placed perpendicular to a uniform magnetic field of 0.50 T. Due to external forces, the area of the lo
pogonyaev

Answer:

6.3\cdot 10^{-4} V

Explanation:

According to Faraday-Newmann-Lenz, the induced emf in the loop is given by:

\epsilon=-\frac{\Delta \Phi}{\Delta t} (1)

where \frac{\Delta \Phi}{\Delta t} is the rate of variation of the magnetic flux through the loop.

We know that the magnetic flux through the loop is given by

\Phi = BA

where B is the magnetic field and A is the area of the loop. Since the magnetic field is constant, we can write the variation of flux as

\Delta \Phi = B \Delta A

So eq.(1) becomes

\epsilon=-B\frac{\Delta A}{\Delta t}

and the problem gives us:

B=0.50 T is the magnetic field

\frac{\Delta \Phi}{\Delta t}=-1.26\cdot 10^{-3} m^2/s is the rate at which the area changes

Substituting into the equation, we find

\epsilon=-(0.50 T)(-1.26\cdot 10^{-3} m^2/s)=6.3\cdot 10^{-4} V=0.63 mV

8 0
3 years ago
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