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Schach [20]
3 years ago
12

What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)

Physics
1 answer:
RSB [31]3 years ago
6 0

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

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A forklift pushes a box with a force of 500 N across the floor for a distance of 5.0 m, then turns around and pushes with the sa
yanalaym [24]

Answer:

5,000J

Explanation:

Work = Force x Distance

Distance back and forth is canceled out, so either the answer is + or -

5.0m + 5.0m = 10.0m

500N x 10.0m = 5,000J

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How strong is naruto i need a in depth answer going off his AP and DC i prefer powerscalers since they do this but anyone can an
marin [14]

Answer:

Naruto is kinda strong

Explanation:

If were just talking about just him I think that he is strong for letting go of his past and moving along with his life. Bu t he also isn't strong if you think about it if Naruto didn't have the nine tails he wouldn't be special so he also is not strong there for his only power really comes from the nine tails.

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3 years ago
While anchored in the middle of a lake, you count exactly five waves hitting your boat every 10 s. you raise anchor and start mo
777dan777 [17]
<span>Every 10s 5 waves; t1 = 2s for each wave
 When v = 1.5m/s, 3 waves in 10s t2 = 10 / 3s
  Calculating the frequency in first case f1 = 5 / 10 = 0.5
 Calculating the frequency in second case f2 = 3 / 10 = 0.3
 Using the Doppler formula f = (1-v/c) f0
  For the formula f = f2, v = velocity of boat= 1.5 m/s, f0 = f1, c is velocity of wave 0.3 = 0.5 x (1 - 1.5/c) => 1.5/c = 1 - 0.6 => 1.5/c = 0.4 => c = 1.5/0.4 Velocity of the wave = 3.75 m/s</span>
7 0
3 years ago
A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   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.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate time when displacement = 1.5 meter.

 1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 m/s^2

Substituting

     -1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0

In case of horizontal motion

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    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation 4.9t^2-0.122Vt-1.5=0

    We will get 4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

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Flura [38]

Answer:

A,B,C,D, and F are correct

Explanation:

5 0
3 years ago
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