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bulgar [2K]
2 years ago
5

A pendulum has a mass of 3 kg and is lifted to a height of 0.3 m. What is the maximum speed of the pendulum

Physics
1 answer:
Zepler [3.9K]2 years ago
8 0
The Energy of the pendulum is

mass*gravity*heigh--> 3kg*0.3m*9.81N/kg=8.829

And so the maximum speed is :

Energy=0.5*mass*speed*speed

reshape to speed:

Speed = 2.426 meters per secound
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An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?
Norma-Jean [14]
Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s
6 0
2 years ago
I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

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3 years ago
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Aloiza [94]

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

Condition 1: The stars rise and set perpendicular to the horizon

The observer is at the equator

Condition 2: The stars circle the sky parallel to the horizon

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Condition 3: The celestial equator passes through the zenith

The observer is at the equator

Condition 4: In the course of a year, all stars are visible

The observer is at the equator

Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)

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What is the sequence of energy transformations associated with a hydroelectric dam?
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Gravitational potential energy -> Kinetic energy -> Mechanical energy -> Electrical energy.

The water starts up (potential) and flows down (kinetic), the flowing water turns a big wheel (mechanical) which creates electricity (electrical).

3 0
3 years ago
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It takes Mars about ____ Earth years to orbit the Sun.
OlgaM077 [116]
<span>At this distance, and with an orbital speed of 24.077 km/s, Mars takes 686.971 Earth days, the equivalent of 1.88 Earth years, to complete a orbit around the Sun. This eccentricity is one of the most pronounced in the Solar System, with only Mercury having a greater one (0.205). 

686.971 rounds to 687
HOPE I HELPED!</span>
7 0
3 years ago
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