All categories

3 years ago

In which direction does the electric field point at a position directly north of a positive charge?

Physics

2 answers:

UNO [17]3 years ago

7 0

Answer:

South

Explanation:

navik [9.2K]3 years ago

5 0

Answer:

B. South

Explanation:

An electric field can be defined as the amount of electric force per unit charge. The direction of the electric field can be determined by the motion of a positive test charge under the electric force.

The direction of electric field is radially outward for a positive charge and radially inward for a negative charge. Thus, for the electric field points toward SOUTH at a position directly south of a positive charge.

You might be interested in

Is this right at all

Alja [10]

It is correct! good job :)

7 0

3 years ago

A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh

dlinn [17]

Answer:

a) t = H/v0

b) H = -(v0)²/g

Explanation:

Hi there!

a)The position of the balls can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t.

y0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity.

t = time.

For the ball that is thrown upwards, the initial height is zero, then, the equation can be written as follows:

y = v0 · t + 1/2 · g · t²

The second ball is initially at a height H and the initial velocity is zero. The equation of height for the second ball will be:

y = H + 1/2 · g · t²

When the two balls collide, their height is the same. Then, equalizing both equations we can obtain the time at which they collide:

v0 · t + 1/2 · g · t² = H + 1/2 · g · t²

v0 · t = H

t = H/v0

b) When the first ball is at the highest point its velocity is zero. Using the equation of velocity we can find the time at which the ball is at that point. The equation of velocity is the following:

v = v0 + g · t

At the highest point v = 0.

0 = v0 + g · t

Solving for t:

-v0/g = t

The time at which the first ball is at the highest point is t = -v0/g

The time at which both balls collide was calculated above:

t = H/v0

Then, equalizing both times and solving for H:

H/v0 = -v0/g

H = -v0/g · v0

H = -(v0)²/g

3 0

3 years ago

What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 11.0 m and the downward accel

8090 [49]

<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration. We can use a simple equation to find centripetal acceleration. a = v^2 / r We can use this same equation to find the speed of the car. v^2 = a * r v = sqrt { a * r } v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) } v = 12.7 m/s The speed of the roller coaster is 12.7 m/s</span>

3 0

3 years ago

Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5

andrezito [222]

Answer:

R = 1.2295 10⁵ m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

how angles are measured in radians

θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

$\frac{y}{R} = 1.22 \frac{ \lambda}{D}$