Answer:
For a relative frequency distribution, relative frequency is computed as the class frequency divided by the number of observations.
Answer:
Sarah is right
Explanation:
This is an exercise that differentiates between scalars and vectors.
A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.
In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero
consequently Sarah is right
Answer:
<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>
Explanation:
<u>Operations With Vectors</u>
Given two vectors in rectangular components:

The sum of the vectors is:

The difference between the vectors is:

The magnitude of
is:

The angle
makes with the horizontal positive direction is:

The question provides the vectors:



Calculate:

The magnitude of
is:

The angle is calculated by:

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.
The magnitude of vector d is 16 and the angle with the x-axis is 270°
Answer:
E) d/sqrt2
Explanation:
The initial electric force between the two charge is given by:

where
k is the Coulomb's constant
q1, q2 are the two charges
d is the separation between the two charges
We can also rewrite it as

So if we want to make the force F twice as strong,
F' = 2F
the new distance between the charges would be

so the correct option is E.
The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.