_________ changes from day to day, while ____________ changes from region to region. A) climate, weather B) weather, climate C)
2 answers:
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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:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
