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kotykmax [81]
3 years ago
12

Why is perpetual energy possible?

Physics
1 answer:
yulyashka [42]3 years ago
5 0
Because this energy is everlasting.
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You find a micrometer (a tool used to
BaLLatris [955]

Explanation:

If the micrometer is still in the working condition, it will be hard for the observer to find its accuracy or precision, which could be matched with the meter stick. Its accuracy can be compromised in comparison with the undamaged micrometer, but it would be more chances of highly precise.

The error which is find in the damaged micrometer would be less than as compared to the width of the hash marks on the stick.

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3 years ago
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write a conclusion to the selectivity in which you completely and intelligently describe the characteristics of an object that i
Karo-lina-s [1.5K]

Answer:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle.

Explanation:

6 0
3 years ago
An electric hoist does 56,447 J of work in raising 115 kg load. How high (in meters) was the load lifted?
Ugo [173]

\\ \rm\rightarrowtail W=mgh

\\ \rm\rightarrowtail 56447=1150h

\\ \rm\rightarrowtail h=56447/11150

\\ \rm\rightarrowtail h=49.1m

8 0
3 years ago
Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in
Radda [10]

Answer:

The value is   t =  14.129 \  minutes    

Explanation:

From the question we are told that

  The distance of planet Tatoone is  d =  1.7 \ AU  =  1.7 *1.496* 10^{11}=2.543*10^{11} \ m

   The  speed of light is  c =  3.0*10^{8} \  m/  s

Generally the time taken is mathematically represented as

     t =  \frac{d}{c}

=> t =  \frac{2.543*10^{11}}{3.0*10^{8} }

=>    t =  847.7 \  s

Now converting to minutes

       t =  \frac{847.7}{60}

   =>     t =  14.129 \  minutes    

8 0
3 years ago
A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C
MAVERICK [17]

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

5 0
3 years ago
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