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neonofarm [45]
3 years ago
6

While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega

tively charged rod is removed. What would happen to the vane?
Physics
1 answer:
Paraphin [41]3 years ago
4 0

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

 

   Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

   In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

   If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

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3.278*10^6 I think. Sorry if it’s wrong.
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3 years ago
Mary takes 6.0 seconds to run up a flight of stairs that is 102 meters long. If Mary's weight is 87 newtons, what power has Mary
fredd [130]

Answer: 1479watts

Explanation:

Power is defined as the energy expended or work done in a specific time.

Mathematically,

Power = Workdone/time taken

Since work done is force × distance

Power = force × distance/time

Force = Mary's weight = 87N

Distance = height of the flight = 102meters

Time = 6.0seconds

Substituting in the formula we have;

Power = 87 × 102/6

Power = 1,479watts

Note that the time must be in seconds before usage. If its given in minutes, you will have to convert to seconds

8 0
3 years ago
What are particles of carbon called?
Ainat [17]
Co2 or greenhouse gas
8 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

4 0
3 years ago
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