Answer:
VB − VA = g tAB & (VA + VB)/2 = h / tAB
Explanation:
s = h = Displacement
tAB = t = Time taken
VA = u = Initial velocity
VB = v = Final velocity
a = g = Acceleration due to gravity = 9.8 m/s²
Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used
Answer:
C) No work is required to move the negative charge from point A to point B.
Explanation:
An equipotential surface is defined as a surface connecting all the points at the same potential.
Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.
The work done when moving a charge is given by
where
q is the charge
is the potential difference between the initial and final point of motion of the charge
However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so
And so, the work done is also zero.
Answer:
T= 8.061N*m
Explanation:
The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:
T = Fr
where F is the force, r the radius.
if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.
Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:
r =
r = 0.5374m
Finally, replacing the value of r and F, we get that the maximun torque is:
T = 15N(0.5374m)
T= 8.061N*m
<u>Answer</u>
Mean
<u>Explanation</u>
For any set of date there is a typical value for the probability distribution called central tendency.
There three types of central tendency<em> mean, median and mode.</em>
Mean is the average value of all data set in a distribution.
median is the the central value when the data is arranged in ascending or descending order.
Mode is the value that occurs most in a set of data.
Answer:
a) 4.98m/s²
b) 481.66N
Explanation:
a) Using the Newtons second law of motion
m is the mass of the object
g is the acceleration due to gravity
Fm is the moving force acting along the plane
Ff is the frictional force opposing the moving froce
a is the acceleration of the skier
Given
m = 60kg
g = 9.8m/s²
= 35°
Ff = 38.5N
Required
acceleration of the skier a
Substituting into the formula;
Hence the acceleration of the skier is 4.98m/s²
b) The normal force on the skier is expressed as;
N = Wcosθ
N = mgcosθ
N = 60(9.8)cos 35°
N = 588cos 35°
N = 481.66N
Hence the normal force on the skier is 481.66N