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OleMash [197]
3 years ago
8

How does an electroscope detect negative static electricity

Physics
1 answer:
MA_775_DIABLO [31]3 years ago
5 0
<span> An electroscope that might have a static charge is tested on a metal surface. From there, the charges move to the metal and straight to the foil leaves. If they repel, or move away from each other, that means they have identical charges. This applies for both positive and negative static electricity.

work cited: School of champions, Google
</span>
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A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag
olga2289 [7]
Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.
8 0
3 years ago
If A = 5 and C = 13, what is sin x?
lisabon 2012 [21]
<span>By pythagorean theorem then, the vertical side of the right triangle must be 12.
Then if x is the angle between the horizontal side and the hypotenuse, sin(x) = 12/13 but also the anser should be in this sentences.
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6 0
3 years ago
When cleaning up a local beach, students found many different particles that were in the water that affected the shoreline, like
Wewaii [24]

The correct answer is D,

4 0
3 years ago
2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe
ValentinkaMS [17]

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

t = \frac{2vsin \theta}{g}

t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

5 0
3 years ago
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
Korolek [52]

Answer:

a

    z= 2.5 \ m

b

   z =  (1 \ m ,  4 \ m )

Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

      z =  4\ m

and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

7 0
3 years ago
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