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Free_Kalibri [48]
3 years ago
6

A lighthouse is located on a small island 5 km away from the nearest point p on a straight shoreline and its light makes two rev

olutions per minute. how fast is the beam of light moving along the shoreline when it is 1 km from p? (round your answer to one decimal place.)
Physics
1 answer:
lisabon 2012 [21]3 years ago
7 0

The distance starting from the point to the lighthouse would be regarded as the hypotenuse.

And also will be the radius of the circle the beam of light is generating at that point. 


So get the radius first

r = sqrt (1^2 + 5^2)

r = 5.099 km


find the circumference:

C = 2*pi * 5.099 km

C = 2 * 16.01898094

C = 32.04 km


Then find the speed in km/sec

One revolution: 60/2 = 30 sec per revolution

Speed = 32.04 km/30 sec

S = 1.068 km/sec is the speed of light

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If you add 1 proton to Carbon, it will no longer be Carbon, it will be ________________.
Whitepunk [10]

Nitrogen

Explanation:

Adding one proton to a carbon atom makes Nitrogen.

A quick introspection on atoms:

  • An atom is made up of three fundamental particles.
  • They are protons, neutrons and electrons.
  • The protons are positively charged and the neutrons do not carry any charges.
  • Electrons are negatively charged.

The difference between an atom and another is the number of protons in them. This is the atomic number.

The periodic table of element is a list of elements arranged based on the number of protons they have. Every element on the table has unique number of protons which makes it differ from another.

  • Atoms do not readily lose their protons because they are held by nuclear forces in the nucleus of an atom.

When an element gains a proton, it becomes another element.

    Carbon has proton number of 6

 If a proton is added to it, it becomes 7

This is the proton or atomic number of nitrogen.

Learn more:

Atomic number brainly.com/question/5425825

#learnwithBrainly

4 0
3 years ago
Answer them all please!!
liubo4ka [24]

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6 0
2 years ago
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
3 years ago
What is the volume of a cube with length = 3 centimeters, width = 3
suter [353]

Answer:

D. 27 cubic centimeters

Explanation:

l*w*h

3*3*3= 27

4 0
2 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0
3 years ago
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