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mylen [45]
3 years ago
11

Help please I need this

Physics
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

yes C would be technically correct with the information given.

Explanation:

If you have any questions feel free to ask in the comments - Mark

Also when you have the chance please mark me brainliest.

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How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?
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There is too much information given, it's hard to understand exactly which variables are important in this problem. 
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3 years ago
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
Harrizon [31]

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;

W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}

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

F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}

where;

  • M is the mass of Earth = 5.972 x 10²⁴ kg
  • R is the Radius of Earth = 6,371 km

For Planet Tehar;

g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g

For planet Loput:

g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g

For planet Cremury:

g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g

For Planet Suven:

g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g

For Planet Pentune;

g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g

For Planet Rams;

g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g

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

W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb

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

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Kinetic Energy - What does it depend on?
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Answer:

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Explanation:

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Fred has a very detailed model of the solar system where each planet is made out of granite rock. Since his little sister really
eimsori [14]
C or "Density" The granite rock has a much higher density than the Styrofoam. 
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3 years ago
Read 2 more answers
Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ
Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}

Put the value into the formula

525=\sqrt{\dfrac{c+v}{c-v}}\times950

\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2

\dfrac{c+v}{c-v}=0.305

c+v=0.305\times(c-v)

v(1+0.305)=c(0.305-1)

v=\dfrac{0.305-1}{1+0.305}c

v=−0.532c

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0
3 years ago
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