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DENIUS [597]
3 years ago
5

Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

, 2, 1, 4 D. 3, 1, 2, 4

Physics
1 answer:
sweet [91]3 years ago
4 0

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

E\propto\dfrac{q}{r^2}

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

E=\dfrac{kq}{r^2}

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

E=\dfrac{k2q}{(2r)^2}

E=\dfrac{kq}{2r^2}

The electric field will be smallest.

(III).  The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

E=\dfrac{k2q}{(r)^2}

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

E=\dfrac{kq}{(2r)^2}

E=\dfrac{kq}{4r^2}

The electric field will be very smallest.

So, The electric field from largest to smallest will be

E_{3}>E_{1}>E_{2}>E_{4}

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

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From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
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Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

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8 0
1 year ago
Two families meet at the park at 10:00am. Each family uses a different way of transportation to get there. The Gonzalez family l
pav-90 [236]

Firstly, we need to make the times and distance equal to compare.

1hr=60mins
Gonz drove 90km in 60mins,so they would drive 45km in 30 mins dividing it by 2.
Because the Rivs had gone 30km in 30mins, Gonz were faster.

For the adverage speed, we should first add the speed of each family in mph , so 90+60, which equals 150 and divide that by 2 because there are 2 speeds so the average speed is 75mph.

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What effects a materials resistance
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The thickness of the resistor.

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8 0
3 years ago
A ping pong ball with a dent in it can be put into a pan of boiling water. After a short amount of time, the dent will pop out.
kifflom [539]

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6 0
4 years ago
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A 65.8-kg person throws a 0.0413 kg snowball forward with a ground speed of 32.5 m/s. A second person, with a mass of 58.7 kg, c
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Answer:

v_{1} = 2.490\,\frac{m}{s}

v_{2} = 0.023\,\frac{m}{s}

Explanation:

The statement is described physically by means of the Principle of Momentum Conservation. Let assume that first person moves in the positive direction:

First Person

(65.8\,kg)\cdot (2.51\,\frac{m}{s}) = (65.8\,kg)\cdot v_{1} + (0.0413\,kg)\cdot (32.5\,\frac{m}{s} )

Second Person

(0.0413\,kg)\cdot (32.5\,\frac{m}{s})+(58.7\,kg)\cdot (0\,\frac{m}{s})=(0.0413\,kg+58.7\,kg)\cdot v_{2}

The final velocities of the two people after the snowball is exchanged is:

v_{1} = 2.490\,\frac{m}{s}

v_{2} = 0.023\,\frac{m}{s}

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