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Anarel [89]
3 years ago
9

You're the first person to visit Mars and you've just met a group of Martian schoolchildren. They are playing street hockey with

parts from the Mars Polar Lander and, after years of watching Star Trek reruns, they are fluent in English. One of them asks you how your weight and mass have changed since you left Earth. You take a moment to measure both and reply correctly that
(A) your weight is essentially unchanged but your mass is less than on Earth.
(B) your mass is essentially unchanged but your weight is less than on Earth.
(C) your weight and mass have both changed significantly.
(D) neither your weight nor your mass have changed much.
Physics
1 answer:
Liula [17]3 years ago
8 0

Answer: (B) your mass is essentially unchanged but your weight is less than on Earth.

Explanation:

Well, firstly we have to explain the difference between the concepts of mass and weight:

<u>-Mass</u> m is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and <u>remains the same regardless of where the body is located.  </u>

<u>-Weight</u> W is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass of the body by the acceleration due gravity g in that place and is mathematically expressed as:  

W=m.g  

So, according to this explanation and since the Earth and Mars have different values ​​of acceleration due gravity (in fact, Mar's g is less than Earth's), <u>your weight has changed, but not your mass.</u>

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A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s
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Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m

The spring constant of the net is 20130.76 N

From Hooke's Law

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m

The net would strech 0.03167 m

If h = 35 m

From energy conservation

65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0

Solving the above equation we get

x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45

The compression of the net is 1.52 m

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3 years ago
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