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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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(The answer is probably obvious for some but I'm not big on science. Not my cup of English-..I'm implying Enlish is more my styl
Lelechka [254]
D because it do da skiddly bip bop boop
4 0
2 years ago
An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12
kati45 [8]

Answer:

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} +  \frac{1}{q}

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

         \frac{1}{q} = \frac{1}{f} - \frac{1}{p}

     

we calculate

          \frac{1}{q} = - \frac{1}{12} - \frac{1}{29}

          \frac{1}{q} = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

          m = \frac{h'}{h} = - \frac{q}{p}

       

we substitute

          m = - \frac{-8.488}{29}

          m = 0.29

the positive sign indicates that the image is right

4 0
3 years ago
Please help me find out this answer
Elanso [62]
The direction of work.........

4 0
3 years ago
Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball
vlabodo [156]

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

6 0
3 years ago
10. According to Newton's First Lawy , if a box is pushed with no external resistance, what will the action of the box be ?
kati45 [8]

Explanation:

According to Newton's First Law of motion, if a box is pushed with no external resistance, the box will keep on moving due to the absence of external force. It might gets stopped due to frictional force that is acting between the surface and the ball. The first law of motion is also known as law of inertia. the magnitude of force acting on the object is given by second law of motion.

4 0
3 years ago
Read 2 more answers
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