Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The capacitor is charged and stores 8.11 × 10 − 9 J of energy. Find the electric field strength E inside the capacitor.

Answers

Answer 1
Answer:

Explanation:

It is known that the relation between electric field and potential is as follows.

             E = (V)/(d)

And, formula to calculate the capacitance is as follows.

           C = (\epsilon_(o) A)/(d)

              = (8.854 * 10^(-12) * (0.479 m)^(2))/(0.479 * 10^(-3))

              = 4.24 * 10^(-9) F

Hence, energy stored in a capacitor is as follows.

         W = (1)/(2)CV^(2)

          V = \sqrt{(2W)/(C)}

        E = \sqrt{(2W)/(d^(2)C)}

            = (2 * 8.11 * 10^(-9) J)/((0.479 * 10^(-3))^(2) * 4.24 * 10^(-9))

            = 16.687 * 10^(3) N/C

Thus, we can conclude that electric field strength E inside the capacitor is 16.687 * 10^(3) N/C.


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Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitely remote ones) does the 72.6 kg mass experience a net force of zero? Answer in units of m

Answers

Answer:

Explanation:

The force due to gravitation is equal to zero for each of the masses.

M1= 181kg

M2= 712kg

m = 72.6kg

The distance between M1 and M2 is said to be fixed , therefore no value should be given I.e it's a constant.

From the formula for gravitational force we have that

F = GMm/r^2

GmM1/(d-r)^2. = GmM2/r^2

where r is the distance between the 72.6 kg and 712kg

d is the distance between M1 and M2

Solving mathematically

r(√M1+√M2) = d√M2

r = d√M2/√M1 + √M2

d×26.68/ 13.45+26.68

d×26.68/40.13

r = 0.665d

A car traveling at 45 km/h starts to brake, and comes to a stop over a distance of 18 m. Calculate the accelerationof the braking car.

Answers

Answer:

Acceleration, a=8.68\ m/s^2

Explanation:

Given that,

Initial speed of a car, u = 45 km/h = 12.5 m/s

Final speed, v = 0 (as they comes to rest)

Distance, d = 18 m

We need to find the acceleration of the breaking car. Using third equation of motion as follows :

v^2-u^2=2ad\n\n\text{Where a is acceleration of the car}\n\na=(v^2-u^2)/(d)\n\na=((12.5)^2)/(18)\n\na=8.68\ m/s^2

So, the acceleration of the braking car is 8.68\ m/s^2.

1. On a force vs. mass graph, what would be the slope of the line?2. On a Free Body Diagram, if the forces are all balanced, what do you know about the
object? Can it be moving?

Answers

1. By Newton's second law,

F = ma

so the slope of the line would represent the mass of the object.

2. If all the forces are balanced, then the object is in equilibrium with zero net force, which in turn means the object is not accelerating. So the object is either motionless or moving at a constant speed.

Final answer:

The slope on a Force vs. Mass graph represents acceleration. In a Free Body Diagram, if all the forces are balanced, the object could be either at rest or moving at a constant velocity.

Explanation:

1. On a Force vs. Mass graph, the slope of the line represents acceleration, according to Newton's second law of motion, which is force equals mass times acceleration (F=ma). The slope of the line is calculated as the change in force divided by the change in mass, which results in acceleration.

2. In a Free Body Diagram, if all the forces are balanced, it means the net force acting on the object is zero. This does not necessarily mean that the object is stationary. The object could be at rest, or it could be moving at a constant velocity. If an object is moving at a constant velocity, it is said to be in equilibrium because the forces are balanced.

Learn more about the Physics of Forces and Motion here:

brainly.com/question/36922560

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An earthquake on the ocean floor produced a giant wave called a tsunami. The tsunami traveled through the ocean and hit a remote island, causing a lot of damage. Is the water that hit the island the same water that was above the earthquake on the ocean floor?A No, the water from above the earthquake stayed in the same place and only the energy was transferred.
B No, the energy in the wave pushed the water particles from above the earthquake in the opposite direction.
C Yes, the water particles moved toward the island while the energy remained above the earthquake.

Answers

A lot of the biology particles can be certified by a sphychiatrist that can determine weather they are infected with Ebola

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.

Answers

Answer:

B

Explanation:

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) =  514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ  / d  ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w =  ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

   = 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations  : brainly.com/question/4449144

Answer:

1.7* 10^(-3) m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

w = (D\lambda )/(d)

w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))

w = 1.7* 10^(-3) m