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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.)

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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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

$a = \frac{vf - vi}{t}$

=

$\frac{15 - 25}{5}$

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0

3 years ago

Explain why the top of the loop cannot be the same height as (or higher than) the top of the first hill. Assume the roller coast

Ivahew [28]

Answer:

By conservation of energy, it can climb up to a height equal to that it went down before. However, due to the friction in the machines, the total mechanical energy of the roller coaster will decrease. As a result, the first "hill" of many roller coasters are the highest, but the followings will have decreasing heights.

Explanation:

7 0

3 years ago

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Damm [24]

Material medium electric waves

7 0

2 years ago

Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en

Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0

3 years ago