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:
4.5 s, 324 ft
Explanation:
The object is projected upward with an initial velocity of
The equation that describes its height at time t is
(1)
where t, the time, is measured in seconds.
In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):
(2)
At the maximum heigth, the vertical velocity is zero:
v(t) = 0
Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:
And by substituting this value into eq.(1), we also find the maximum height:
The moon is between the sun and earth.
The side where the light from the sun hits the moon is facing away from earth.
Answer:
Correct answer: 11. Total distance d = 200m ; 12. Vav = 3.63m/s ;
13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;
15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s
Explanation:
The whole movement can be divided into four stages.
In the first stage the subject moves 60m in a positive direction for 10s,
in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.
11. Total distance = 60 + 0 + 100 + 40 = 200m
12. The formula for calculating the average speed (velocity) is
Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)
Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s
13. The movement started from the origin and ended at the origin
Total displacement is zero meters.
14. The speed between 10s and 15s is zero, because he did not move.
15. V₃ = S₃/t₃ = 100/25 = 4 m/s
16. V₁ = S₁/t₁ = 60/10 = 6 m/s and V₄ = S₄/t₄ = 40/15 = 2.67 m/s
V₁ > V₄
God is with you!!!
