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jasenka [17]
3 years ago
14

Consider two positively charged particles, one of charge q0 (particle 0) fixed at the origin, and another of charge q1 (particle

1) fixed on the y-axis at (0,d1,0). What is the net force F⃗ on particle 0 due to particle 1?
Physics
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

F12= (kq1q2/r12 squared)*e12

Explanation:

F12→ is the force q 1 makes on q 2 and e12 is the unit vector from q1 toward q2.

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PLEASE HELP AS SOON AS POSSIBLE!!
Naily [24]

Answer:

Saved Energy ?

Explanation:

5 0
3 years ago
A 55 kg skater spins 12 m/s while carving a circle on the ice that has a radius of 6.0m. What net force must act on the skater t
vivado [14]
F = m.a
a = v^2 / r
a = 12^2 / 6.0
a = 24 m/s^2
F = 55 × 24
F = 1320 N
4 0
3 years ago
When you look at the backside of a shiny teaspoon held at arm's length, do you see yourself upright or upside down? (b) When you
igor_vitrenko [27]

Answer:

a) The back spoon gives a right image (upright)

b) the front gives an inverted image

Explanation:

The spoon is a curved metallic object, when we see ourselves from the back we have a convex mirror, in this type of mirror when the law of reflection is applied the rays diverge therefore the eye-brain system forms the image with the prolongation of the rays, therefore the image is straight and smaller than the object.

When we look through the deep side of the spoon, we have a concave mirror and as the object (we) is further away than the distance, the rays converge to a point, so the image is real, inverted smaller than the object.

In summary.

a) The back spoon gives a right image (upright)

b) the front gives an inverted image

3 0
3 years ago
An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e
zimovet [89]

Answer:

a) v = 6.25\times 10^{5} m/s

b) v = 1.73\times 10^{4} m/s

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is 4.55\times 10^5 m/s

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

a =\frac{q E}{m}

a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}

a = 2.58\times 10^{11} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

      = 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355

v = 6.25\times 10^{5} m/s

b) for proton

a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}

a = -1.41\times 10^{8} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

       = 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355

v = 1.73\times 10^{4} m/s

3 0
3 years ago
Two students are standing on a fire escape, one twice as high as the other. Simultaneously, each drops a ball.
almond37 [142]
The second ball should strike at double the original t value
4 0
3 years ago
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