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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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In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a
MaRussiya [10]

Answer:

v= \frac{10\times10^{-3} V}{70} m/s

Explanation:

Assumption: bullet leaves the muzzle at a speed of V m/s

and velocity of push received by the man be v m/s

According to newton's third law to every action there is always an equal and opposite reaction.

therefore,

mass of man× velocity = mass of bullet×its velocity

⇒70×v= 10×10^-3 ×V

solving the above eqaution we get

therefore v= \frac{10\times10^{-3} V}{70} m/s

7 0
4 years ago
A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b
Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 m/s^2

So 4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}

Now at time 2t,

  S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g m/s^2

4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}

At time 2t,

     S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm

 So vertical distance traveled in time 2t = 16 cm to the bottom

 

6 0
3 years ago
On a day the air is still a windmill still posses what type of energy is that ?
Montano1993 [528]

Well, it's up on top of a pole or pedestal of some sort,
so it has some gravitational potential energy relative to
the ground.  In other words, if it somehow became detached
from its structure and fell to the ground, it would make quite
an energetic splat when it got there.

Also, the windmill is at the temperature of the air around it,
which is far from Absolute Zero, so the windmill holds a lot of
thermal (heat) energy.

Then I guess there's the matter of the chemical energy in the
molecules of the material that the windmill is made of, and the
nuclear energy in its atoms.

6 0
3 years ago
Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts hi
Sati [7]

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

 

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

8 0
4 years ago
Will give brainliest if they answer is correct ​
My name is Ann [436]

Answer:

When a motorcycle takes a turn, centrifugal force—in this case, friction between the tires and the road—pushes it towards the center. This basic physics explains why riders can lean into turns without falling. However, when an outside force disrupts or unbalances these forces, the vehicle crashes. that is the only one that I can answer for you. :)

6 0
3 years ago
Read 2 more answers
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