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3 years ago

Which statement is the correct representation of these electric field lines?

Physics

2 answers:

irinina [24]3 years ago

6 0

C . plate a is negatively charged and plate b is positively charged

Taya2010 [7]3 years ago

5 0

Answer:

C. Plate A is negatively charged and plate B is positively charged.

Explanation:

The direction of the lines of an electric field represents the direction to which a positive test charge placed in that field would move. Therefore, since positive charges are repelled by positive charges and attracted by negative charges, this means that:

- Electric field lines point away from positive charges, and

- Electric field lines point towards negative charges

In the figure in the problem, we see that electric field lines go away from plate B and towards plate A. This means that:

- Plate B is positively charged

- Plate A is negatively charged

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2. A ball tied to a pole by a rope swings in a circular path with a centripetal acceleration of 2.7 m/s. If the ball has a

Helga [31]

Answer: The diameter of the circular path is 2.96m

Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.

Centripetal acceleration = 2.7m/s^2

Tangential speed = 2.0m/s

Radius = 2.0^2 / 2.7 = 4/2.7

= 1.48m

Diameter = radius*2

= 1.48*2 = 2.96m.

3 0

3 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

1 year ago

B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

prohojiy [21]

I Lolo my name is keshav and

5 0

3 years ago

Hurricanes are especially destructive because of the abrupt decrease in air pressure that they bring with the winds. Consider a

ivann1987 [24]

Answer:

2.96 × 10^4 N

Explanation:

1 atm = 101325 N/m², pressure inside the airtight room = 1.02 atm, pressure outside due to hurricane = 0.91 atm

net pressure directed outward = P inside - P outside

net pressure = 1.02 - 0.91 = 0.11 atm

where 1 atm = 101325N/m²

0.11 atm = 0.11 × 101325 N/m² = 11145.75 N/m²

area of the square wall = l × l where l is the length of the wall in meters = 1.63 × 1.63 = 2.6569

net pressure = net force / area

make net force subject of the formula

net force = net pressure × area = 11145.75 × 2.6569 = 2.96 × 10 ^4 N

8 0

3 years ago

Without steering, a car (mass 1501 kg) can travel with a certain speed around a banked frictionless corner angled at 25.0° to th

natima [27]

Answer:

Its tricky but am still working on it

6 0

3 years ago