Your average speed is 20 mph, but ONLY IF YOU'RE IN THE CAR during those 2 hours.
Answer:
Shawn's kinetic energy is 61.45 J.
Explanation:
Given;
Mass of the system is, 
Displacement of the bike is, 
Time taken is, 
Let the constant velocity be
.
For constant velocity, magnitude of velocity is given as distance by time.
Therefore, 
Now, kinetic energy of a body is given as:

Here, 

Therefore, Shawn's kinetic energy is 61.45 J.
Answer:
Explanation:
For simple pendulum the formula is

Where T is time period , l is length and g is acceleration due to gravity .

n is frequency
Putting the values


l = 1.584 m
The planet that Punch should travel to in order to weigh 118 lb is Pentune.
<h3 /><h3 /><h3>The given parameters:</h3>
- Weight of Punch on Earth = 236 lb
- Desired weight = 118 lb
The mass of Punch will be constant in every planet;

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

where;
- M is the mass of Earth = 5.972 x 10²⁴ kg
- R is the Radius of Earth = 6,371 km
For Planet Tehar;

For planet Loput:

For planet Cremury:

For Planet Suven:

For Planet Pentune;

For Planet Rams;

The weight Punch on Each Planet at a constant mass is calculated as follows;

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.
<u>The </u><u>complete question</u><u> is below</u>:
Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.
Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).
<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>
Learn more about effect of gravity on weight here: brainly.com/question/3908593
Answer:
Strike-slip fault
Explanation:
Transform boundaries play the role of connecting the other plate boundary segments.
When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.
Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement releases energy and takes place in a horizontal direction which is parallel to the force of compression.