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irina1246 [14]
3 years ago
14

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm

. Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
Physics
1 answer:
mihalych1998 [28]3 years ago
6 0

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?

Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.  

(a) What is the electric potential at point a due to q1 and q2?  

(b) What is the electric potential at point b?

(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer:

a) the potential is zero at the center .

Explanation:

a) since the two equal-magnitude and oppositely charged particles are equidistant

b)(b) Electric potential at point b, v = Σ kQ/r

r = 5cm = 0.05m

k = 8.99*10^9 N·m²/C²

Q = -2 microcoulomb

v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m

v =  -105 324 V

c)workdone = charge * potential

work = -6.00µC * -105324V

work = 0.632 J

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jonny [76]

Answer:

The field of view is reduced.

Explanation:

Given that,

The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.

We know that,

Field of view :

When the view is observed at a point in a defined field then these field called field of view.

The normal angle of  field of view is 90°.

The formula of field of view is define as,  

field\ of\ view = \dfrac{field\ number}{magnification}

We can say that,

The field of view is inversely proportional to the magnification.

When magnification is low then field of view will be large.

When magnification is higher then field of view will be small .

According to question,  

When the magnification adjust from 4 to 10 to 43, the field of view is reduced.    

Hence, The field of view is reduced.

7 0
2 years ago
2. A rock is dropped off a bridge. How fast is the rock
zhenek [66]

Answer:

a) The velocity of rock at 1 second, v = 9.8 m/s

b) The velocity of rock at 3 second,  v = 29.4 m/s

c) The velocity of rock at 5.5 second,  v = 53.9 m/s

Explanation:

Given data,

The rock is dropped from a bridge.

The initial velocity of the rock, u = 0

a) The velocity of rock at 1 second,

   Using the first equation of motion

                         v = u + gt

                         v = 0 + 9.8 x 1

                          v = 9.8 m/s

b) The velocity of rock at 3 second,

                         v = u + gt

                         v = 0 + 9.8 x 3

                          v = 29.4 m/s

c) The velocity of rock at 5.5 second,

                         v = u + gt

                         v = 0 + 9.8 x 5.5

                         v = 53.9 m/s

5 0
3 years ago
57. Estimate Potential Energy A boulder with a
Oksanka [162]

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

3 0
3 years ago
A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th
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Answer:

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Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

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The final carriage speed is zero (vf = 0)

      0 = v₀² - 2ax

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      a = 3.025 m / s²

      a = 3.0 m/s²

We calculate the average force

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3 0
3 years ago
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Nuetrik [128]

Answer:

Cannot be determined from the given information

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Velocity = 24 m/s

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x = Asin(ωt + ϕ)

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x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.

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