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2 years ago

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

Physics

2 answers:

irinina [24]2 years ago

6 0

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

Taya2010 [7]2 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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14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

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8 0

6 months ago

Find the slit separation of a double-slit arrangement that will produce interference fringes 0.018 rad apart on a distant screen

inessss [21]

Answer:

1.64 * 10^(-5) m

Explanation:

Parameters given:

Angular separation, θ = 0.018 rad

Wavelength, λ = 589 nm = 5.89 * 10^(-7) m

The angular separation when there are 2 slots is given as

θ = λ/2d

where d = separation between slits

d = λ/2θ

d = (589 * 10^(-9))/(2 * 0.018)

d = 1.64 * 10^(-5) m

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2 years ago

How we can best share the observation of the light through one hole the paper

zhenek [66]

Drawing is an easy way to connect to your inner child

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2 years ago

A 2.3 kg cart is rolling across a frictionless, horizontal track towards a 1.5 kg cart that is initially held at rest. The carts

Inga [223]

Answer:

total momentum = 8.42 kgm/s

velocity of the first cart is 3.660 m/s

Explanation:

Given data

mass m1 = 2.3 kg

mass m2 = 1.5 kg

final velocity V2 = 4.9 m/s

final velocity V3 = - 1.9 m/s

to find out

total momentum and velocity of the first cart

solution

we know mass and final velocty

and initial velocity of second cart V1 = 0

so now we can calculate total momentum that is m1 v2 + m2 v2

total momentum = 2.3 ×4.9 + 1.5 ×(-1.9)

total momentum = 8.42 kgm/s

and

conservation of momentum is

m1 V + m2 v1 = m1 v2 + m2 v3

put all value and find V

2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)

V = 8.42 / 2.3

V = 3.660 m/s

so velocity of the first cart is 3.660 m/s

8 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$