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Kisachek [45]
3 years ago
8

A point charge of -2 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m. Find the x and y coord

inates of the position at which an electron would be in equilibrium
Physics
1 answer:
Soloha48 [4]3 years ago
3 0

Answer:

x coordinate = -1.66 m

y coordinate is = -0.825m

Explanation:

Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12

We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2

0.0356s^2 -0.019s-0.0897=0  

s=1.876m

The angle of the line between the two charges is arctan(.5/1) = 26.6o

x coordinate = -1.876*cos(26.6) = -1.66m

y coordinate is -1.876*sin(26.6) = -0.825m

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a roller coaster begins at the top of a hill if it acelerates at the rate of 2 m/s2 and has a mass of 2000 kg what net force is
AlladinOne [14]

Answer:

\boxed {\boxed {\sf 4000 \ Newtons }}

Explanation:

Force can be found by multiplying the mass by the acceleration.

F=m*a

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

m= 2000 \ kg \\a= 2 \ m/s^2

Substitute the values into the formula.

F= 2000 \ kg * 2 \ m/s^2

Multiply.

F= 4000  \ kg*m/s^2

  • 1 kg*m/s² is equal to 1 N
  • Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

F= 4000 \ N

The net force acting on the roller coaster is <u>4000 Newtons.</u>

7 0
3 years ago
During an observation, it was noticed that light diffracts as it passes through small slits in a barrier. What does this evidenc
Lynna [10]

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

8 0
3 years ago
A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h
timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

F_n = \frac{mv^2}{r}

N+mgcos\theta = \frac{mv^2}{r}

Here,

Normal reaction of the ring is N and velocity of the ring is v

N+mgcos\theta = \frac{mv^2}{r}

N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})

N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}

N = 1.374lb

PART B) Acceleration

F_t = ma_t

-mgsin\theta = ma_t

-W sin\theta = \frac{W}{g} a_t

-2Sin30\° = (\frac{2}{32.2})a_t

a_T = -16.10ft/s^2

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

8 0
3 years ago
in the derivation of the time period of a pendulum in electric field when considering the fbd of bob to find the g effective why
Neko [114]

Answer:

we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force that is responsible for the vibration. So what forces act upon a pendulum bob? And what is the restoring force for a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum. The tension force results from the string pulling upon the bob of the pendulum. In our discussion, we will ignore the influence of air resistance - a third force that always opposes the motion of the bob as it swings to and fro. The air resistance force is relatively weak compared to the two dominant forces.

The gravity force is highly predictable; it is always in the same direction (down) and always of the same magnitude - mass*9.8 N/kg. The tension force is considerably less predictable. Both its direction and its magnitude change as the bob swings to and fro. The direction of the tension force is always towards the pivot point. So as the bob swings to the left of its equilibrium position, the tension force is at an angle - directed upwards and to the right. And as the bob swings to the right of its equilibrium position, the tension is directed upwards and to the left. The diagram below depicts the direction of these two forces at five different positions over the course of the pendulum's path.

that's what I know so far

8 0
3 years ago
Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range
Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive  or attractive only

Y: Very small range

Z:  Repulsive and attractive

8 0
3 years ago
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