The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

Here,
= Change in height
m = mass of super heroine
g = Acceleration due to gravity
The change in height will be,

The final position of the heroin is below the ground level,

The initial height will be the zero point of our system of reference,


Replacing all this values we have,



Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J
Answer:
(a) 89 m/s
(b) 11000 N
Explanation:
Note that answers are given to 2 significant figures which is what we have in the values in the question.
(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

(b) The tension,
, is given by

where
is the speed,
is the tension and
is the mass per unit length.
Hence,

To determine
, we need to know the mass of the cable. We use the density formula:

where
is the mass and
is the volume.

If the length is denoted by
, then


The density of steel = 8050 kg/m3
The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is




Answer:
For elliptical orbits: seldom
For circular orbits: always
Explanation:
We start by analzying a circular orbit.
For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.
Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.
So for a circular orbit,
the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.
Now we analyze an elliptical orbit.
An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.
Therefore, for an elliptical orbit,
the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.
Answer:
The vertical velocity of the skater upon landing is 10.788 meters per second.
Explanation:
Skateboarder experiments a parabolic movement. As skateboarder jumps horizontally off the top of the staircase, it means that vertical component of initial velocity is zero and accelerates by gravity, the final vertical speed is calculated by the following expression:

Where:
- Initial vertical speed, measured in meters per second.
- Final vertical speed, measured in meters per second.
- Gravitational acceleration, measured in meters per square second.
- Time, measured in seconds.
Given that
,
and
, the final velocity of the skater upon landing is:


The vertical velocity of the skater upon landing is 10.788 meters per second.
Answer:
The Force between the two charges is an attractive force of 16,000N
Explanation:
Expression for the electric force between the two charges is given by
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
q1 = - 2.0x10^-4C
q2 = + 8.0x10^-4C
r = 0.30 m
Substitute the given values in the above expression -
One charge is + and the other is a -, therefore the net force is an attractive force (opposites atract)
The attraction force is:
F= 9.0x10^9 * 2.0x10^-4 *8.0x10^-4 N/ 0.30^2
F= 16,000N