i-what-is-the-weight-of-a-68-kg-astronaut-a-on-earth-b-on-the-moon-g-17-ms2-c-on-mars-g-37-ms2-d-in-

Name: i what is the weight of a 68 kg astronaut a on earth b on the moon g 17 ms2 c on mars g 37 ms2 d in
Uploaded: 2019-05-10
Duration: 5 min 29 s
Channel: Numerade
Description: (I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?

Question

(I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?

Kathleen Tatem · Accepted Answer

uh, so we&#x27;re going to be solving problem two and checked her for of the gang Coy physics textbook and were given that there&#x27;s a sixty eight kilogram astronaut. We&#x27;re as to find his wage. That&#x27;s the mass of the astronaut. Or has to find his weight on Earth on the moon and on Mars as well as in outer space. Traveling a constant velocity to part a, um, wait on earth we know that way is a force. Maybe Reese thinking about weight is almost just like mass. But masses, um, and inherit property of an object. And the weight is really a force that depends on the acceleration due to gravity. So I&#x27;LL label it W um but this is really force of gravity, usually labeled FT physics textbooks and this sequel to the Mass, just like any force mass times acceleration, except for the celebration, is Teo and on Earth. We know that this is nine point eight approximately nine point eight meters per second squared to play in the mass at the same time sixty eight times bring create, and this will be in new ten&#x27;s kilograms nears per second squared, and we can put that into a calculator. Get six hundred sixty six point four and remember, we only want report. The number of significant digits that we&#x27;re using, which is to significant, is it&#x27;s nine point eight sixty eight. So this is roughly six hundred seventy millions. Some part B for asked to find this fur on the moon where were given that he&#x27;s equal to one point seven communes per second squared. So can we just use the same formula, her weight being Mt. And he just We&#x27;re gonna play these numbers in the calculator. Still the sixty eight kilograms. And now we&#x27;re not going to use nine point eight years for second squared. That&#x27;s exploration due to gravity on Earth. Money&#x27;s the one point seven. If they told us. Given the problem for the acceleration due to gravity on the moon and then approximating Teo to significant digits, you should get one hundred twenty Nunes part See, we&#x27;re just going to do the exact same thing. They also tell us acceleration due to gravity. But on Mars three point seven years per second squared, Yeah, still using the same mask. It&#x27;s all about the same astronaut. Sixty eight kilograms. Oops! And the answer of you should get for that just plugging into calculator approximating Turn fifteen unions and then party his work. It&#x27;s different because we&#x27;re now told that the asteroids in outer space not on a planet, not feeling gravity and he&#x27;s trailing a constant velocity and on velocity is constant. It means that acceleration is zero draw narrow that this implies equal zero because, remember, the acceleration is stuff have lost e changes in time, but it&#x27;s constant. That&#x27;s what we&#x27;re told in the problem. And then wait is a force. This&#x27;s equal to M A. But one a zero means equal to zero. So travelling and constant velocity in outer space, there&#x27;s just no forces acting on the asteroid, so thank you for listening.

Eric Mockensturm · Answer

here we got what is the weight of a asked, what is the weight of a astronaut? That ways that has a mass of 76 kg. And we asked what is the what is um his weight Her weight um on the moon or on Earth? On the moon? On MARS. And then in space with constant velocity, traveling constant velocity. So the weight is just the mass times the gravitational constant. And depending on where they where this astronaut is, the gravitational constant is different. So on Earth the astronaut weighs 746 newtons. Because GS 9.81 they tell us on the moon, G is 17 So the mass times 17 is 129 newtons. So much much lighter on the on the moon than on Earth on MArs, MArs has a much lower gravitational constant than Earth does because it&#x27;s it&#x27;s smaller. Um And so that is the weight on MArs is 281 newtons. So if we ever get an astronaut to MArs, we will way less. And that&#x27;s again, there was an interesting thing recently about the helicopter that they have that can fly on MArs, although the atmosphere is very thin, so it&#x27;s hard to get aerodynamic lift forces. Um The nice thing is, is that the weight that that the lift forces have to overcome is um, is significantly less than it would be on Earth. And then finally, in in outer space, when if a traveling at a constant velocity so that the acceleration of the astronaut is zero, then g is zero and the weight or effectively zero, obviously there&#x27;s gravitational forces from whatever surrounding bodies, but again, those would all be very, very small. Um and so the effect, effectively the weight of a person in in space is just zero and that&#x27;s why we call it weightless.

Matthew Miranda · Answer

question Number 51 asks. What will an astronaut&#x27;s appearance weight be when they&#x27;re in a spaceship 4200 kilometers from the centre of the moon under two different conditions in part A. It stated that the velocity of the spaceship in which the astronaut is traveling is constant. So if we illustrate this, here&#x27;s a surface of the moon down at the bottom. And then right up here is our spaceship, which is housing the astronaut being carried to the moon and inside. I&#x27;ll draw our astronaut now to make it easy on ourselves. Let&#x27;s assume that the astronaut is not moving inside the spaceship there, either seated or strapped in somehow. So there is a normal force being exerted on the astronaut by the spaceship in this direction and subsequently the astronauts wait him. G is acting in the opposite direction. Let&#x27;s use down as the positive direction for this problem. Remember that the spaceship is traveling at a constant velocity. That means that it&#x27;s acceleration is equal to zero acceleration, where criers a change in velocity over a change in time. That also means that the net force acting on the astronaut is also equal to zero. So in taking the some of the forces on the astronaut, we would have m g minus f some n equal to zero. That means that our normal force will be equal to M G and M G. He is going to be equal to the gravitational, constant times the mass of our astronaut times. The mess of the moon, divided by the distance between the astronaut on the moon, squared basically the measure of how much the moon&#x27;s gravity pulls on the astronaut. We know a few values that we can plug in here. We know the mass of our astronaut imps of a is equal to 75 kg. The mass of the moon is equal to 7.35 times 10 to the 22nd kilograms and are the distance between the astronaut on the moon center is equal to 4200 kilometers, which is 4.2 times 10 to the 6 m meters air. The unit we want to use and solving this problem know that we have all our Constance. We can find the astronauts apparent weight, otherwise known as the normal force. EFS of end will be equal to 6.67 times 10 to the minus 11 Newton&#x27;s times meters squared over kilograms squared gravitational constant times. The 75 kg mass of our astronaut times the moon&#x27;s 7.35 times 10 the 22nd kilogram mass divided by 4.2 times tend to the 6 m squared. When we saw this equation, we find that the apparent weight of R S you&#x27;re not is equal to 20.84 mutants. Since this value is positive and be chose down as our positive direction for the problem, the astronauts will feel that their weight is directed downward towards the moon in the second part of our problem. It stated that the astronaut is now accelerating downward towards the moon&#x27;s surface at a rate of 2.9 m per second squared. So let&#x27;s re illustrate this problem to show our current situation. There&#x27;s the moon&#x27;s surface again and now our rocket carrying the astronaut is accelerating downwards towards the moon. I assume that means it&#x27;s getting ready toe land. Our astronaut is inside it right there. Now the normal force of the ship pushing against the astronaut f sub in will still be pointed in the upward direction. Well, the astronauts weight MGI well, we pointed downward but now we have acceleration also pointing downward along with the astronauts. Wait so the some of the forces for this part of the problem will be equal to M G minus efs of n who use the same conditions as the last part of the problem and say that down is positive Onley This time, instead of equaling zero thes forces will be equal to m a mass times acceleration. Now the apparent weight or normal force of our astronauts will be equal toe MGI minus m a. We hunt mg in the last part of our problem that was equal to 20.84 mutants. So we can plug that into our formula for F seven and say that our astronauts apparent weight will be equal toe 20.84 Newtons minus 75 kg times their acceleration rate off 2.9 m per second squared. Calculating this out gives our astronauts apparent weight of value of minus 196.7 Newtons. The fact that it&#x27;s negative means that the weight of the astronaut is now being directed opposite to the positive direction downward. So presumably the astronauts weight is being directed this way because as the spaceship is landing, maybe they&#x27;re being pulled down to their seat by safety belts pushed against the ceiling by the acceleration downward.

Linda Winkler · Answer

here we are going to examine the difference between mass and weight. So we&#x27;re a reminder that mass is the property of a single object and it&#x27;s often associated with the inertia of the object. There&#x27;s also gravitational mass, but it&#x27;s the mass of a single object that interacts through gravity. Wait, on the other hand, is the force of gravity between two objects. So it&#x27;s a force interaction pair. So force of two objects interacting via gravity. And as an example, let&#x27;s take an astronaut who has a weight of 250 newtons on the moon from the units we can tell that weight is of force. And we&#x27;re going to use the surface gravity of the moon, Which is roughly 1/6 of what it is on the earth, 1.63 m/s squared. And we&#x27;re going to deduce the mass of this astronaut. So wait is equal to mass times G. And there we see the interaction taking place. The astronauts mass and the the gravity of the moon are interacting. So we can determine the mass of this astronaut. 250 newtons over 1.63 meters per second squared will give us their mass in kilograms. Uh huh. And that&#x27;s about 153. Not that astronaut comes to the Earth, so that&#x27;s the astronaut on the moon. But if they come back to the Earth, which we hope they do, then that same astronaut will experience await equal to their mass times the G. On Earth. And we can take their mass 153 kg and multiply by the 9.8 meters per second squared, which comes from the Earth and we get something much bigger. Their weight is six times bigger, then it would be on the moon. But their mass Is still 1:53 kilograms. Does not matter if they are on the Earth or mars or in orbit around the Earth, etcetera. As long as they don&#x27;t shed gear or go on a diet, their mass is a property of them and everything that they&#x27;re wearing, it stays the same.

Rachel Wellington · Answer

Hey, everyone, this is question number fifty three from Chapter five. This problem as what is the apparent weight of seventy five kilogram astronaut twenty five hundred kilometres from the centre of the moon in a space vehicle, a moving at constant velocity and be accelerating towards the moon. And we&#x27;re supposed to stay the Russian so of I drew everybody. Diagram F. Ennis pointing up normal force. That is the magnitude of his apparent weight. That&#x27;s what&#x27;s pushing back up. We&#x27;re assuming he&#x27;s in contact with this patient, either stopped into a chair, standing on it, the space vehicle, whatever. But that that space vehicle is exerting an equal and opposite force to his apparent weight. And then we have his his actual way mg, and we&#x27;re going to note down as positive because great, because the moon&#x27;s gravity is going to be pulling him down. Okay, so let&#x27;s start with a some of forces, as always when we&#x27;re dealing with accelerations and apparent weights and stuff. So I&#x27;m jeez, positive minus f n. If his, um moving at a constant glossy, then there&#x27;s no acceleration. That&#x27;s that&#x27;s going to be equal to zero, which gives us F N equals mg and we know when we&#x27;re dealing with a planet. M G is equal to G m m of the object and of the planet over r squared which is the radius the radius of the object to surface of the planet and that is going to equal FN. So now we can solve f n equals G six point six seven times ten to the minus eleven Newton Meet Newton meters square per kilogram squared Time&#x27;s massive the ash, not seventy five kilograms mass of the moon. Seven point four times ten to the twenty two kilograms over the radius squared. This case is from the surface or two point five times ten to the six meters because we&#x27;re told it&#x27;s twenty five hundred kilometers and that&#x27;s squared. And that fifty nine point three so f n equals fifty nine point three. Nine point two three Excuse me, fifty nine point two three noons and because this is positive, the FN value that we got his positive, which means it is pointing in the same direction as we originally did, noted it as so, it&#x27;s fifty nine point two three. Newton&#x27;s away from the moon. We can a process approximate to fifty nine Newtons away from the moon. So that&#x27;s the apparent way at a constant velocity. No, now I&#x27;m going to part B, which is the afternoon is accelerating towards the moon. So now we write again. Second law f equals mg minus fn. Except this time we have an equal to a mass times acceleration because he&#x27;s accelerating toward the moon. So now we have we&#x27;re still solving for FN ad tio sides attract may. So we have FN equals m g minus. We just solved for mg which is fifty nine point two three. So f n equals fifty nine point two three Nunes minus I am a massive the astronaut times his acceleration which were given as one point eight meters per seconds squared and it&#x27;s towards the moon. So that is positive and that equals you get minus seventy six Nunes. So we got f n is equal to minus seventy six, which means this is in the opposite direction is originally denoted originally FN is to notice pointing up which we said his negative away from the moon. So this is seventy six Newton&#x27;s toward the moon because it&#x27;s opposite of that So maybe the national is pinned against the quote ceiling of the spacecraft or safely safety belts. They&#x27;re pulling down something like that when he&#x27;s accelerating, but that&#x27;s why it&#x27;s toward the moon.

Meghan Miholics · Answer

so in discussion were given a weight oven astronauts while standing on Earth. His way is 800 Newtons and were asked to find his weight on Mars, where there is a different acceleration due to gravity of 3.76 meters per seconds Where so this question, um you know, is relying on your knowledge of the difference between weight and mass. So waits refers to the gravitational force F g cry so F g is the weight. And so of course that&#x27;s going to depend on what planet or body you&#x27;re on. Ah, because it depends on the particular value for G that that object has. So the weight of an object on a certain planet will be sometimes g where G is the acceleration due to gravity on that planet. And M is theme mass of that that particular object. So using this equation and the fact that he has or he or she has a weight of 800 Newtons on Earth, we can calculate the mass. Right, So he has 800 Newtons. Ah, wait. And then, um, the acceleration due to gravity on earth is 9.81 so we can solve for the mass of this astronauts as 800 divided by 9.81 When you put that into your calculator, you get a value of 81.55 kilograms. So that&#x27;s the mass of the astronaut. It&#x27;s how much material is actually composing the person or the objects. And then we can use that with the new value for G to calculate the weights on Mars. So I&#x27;ll do that over here. So instead of using a value of 9.81 Fergie, Mars has a different value of G it it&#x27;s 3.76 That&#x27;s what we&#x27;re going to use here. So the weight on Mars will be F g. So that equals the mass that we found in the first calculation times G on Mars. So that will be 81.55 times 3.76 And so we get a weights on Mars of 307 Newtons. So if the astronaut has a weight on Earth of 800 Newtons, the weight on Mars is going to be less. It&#x27;s gonna be about 300 Nunes, because the gravitational pull is not a strong on Mars

Problem 2 Easy Difficulty

(I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?

Answer

a) $670 \mathrm{N}$
b) $120 \mathrm{N}$
c) $250 \mathrm{N}$
d) $0$

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a) $670 \mathrm{N}$b) $120 \mathrm{N}$c) $250 \mathrm{N}$d) $0$

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Andy C.

Liev B.

Marshall S.

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Math Review - Intro

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Andy C.

Liev B.

Marshall S.

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a) $670 \mathrm{N}$
b) $120 \mathrm{N}$
c) $250 \mathrm{N}$
d) $0$

Douglas C. Giancoli

Physics