Answer:
The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Explanation:
Given that,
Point charge at origin = 2 nC
Second charge = 5 nC
Distance at x axis = 8 m
We need to calculate the electric field at the point x = 2 m
Using formula of electric field

Put the value into the formula


The direction is toward positive x- axis.
Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Answer:
<em>d. 268 s</em>
Explanation:
<u>Constant Speed Motion</u>
An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.
Expressed in a simple equation, we have:

Where
v = Speed of the object
d = Distance traveled
t = Time taken to travel d.
From the equation above, we can solve for d:
d = v . t
And we can also solve it for t:

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.
The time it will take for them to meet is:

t = 0.0746 hours
Converting to seconds: 0.0746*3600 = 268.56
The closest answer is d. 268 s
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer:
It comes out the positive side of the battery and goes in to the negative side of the battery
Explanation:
There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.
In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.
However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative
Hi there!
Two possible answers are air resistance and friction.
Friction is caused by the rubbing of the surface of the ground and the surface of the object. Although ice doesn't have much friction, it can still cause friction.
Air resistance is caused by friction between the air and the object. As the object moves along a surface, it collides into many air particles; thus, it slows down.
Hope this helps.
Have an awesome day! :)