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ra1l [238]
3 years ago
8

the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located

2.00 mm above the dipole's midpoint

Physics
1 answer:
vichka [17]3 years ago
3 0

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

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lara [203]

Answer:

P = 33.6 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces [N]

m = mass = 14 [kg]

a = acceleration = 6 [m/s²]

F = 14*6\\F = 84 [N]

In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

W = F*d

where:

W = work [J]

F = force = 84 [N]

d = displaciment = 40 [m]

W = 84*40\\W = 3360 [J]

Finally, the power can be calculated by the relationship between the work performed in a given time interval.

P=W/t\\

where:

P = power [W]

W = work = 3360 [J]

t = time = 100 [s]

Now replacing:

P=3360/100\\P=33.6[W]

The power is given in watts

3 0
2 years ago
A basketball is tossed up into the air, falls freely, and bounces from the wooden floor. From the moment after the player releas
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Answer:

Tha ball- earth/floor system.

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Another way to understand it is:  The distance from which the Earth's orbit appears 1 arcsecond across.

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1 arcsecond is 1/3600 of a degree, 0.00028 degree.  

8 0
3 years ago
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the
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Answer:

4.47 km

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From Pythagoras theorem

AD² = AE² + ED²

AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km

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