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13. A whale comes to the surface to breathe and then dives atan angle of 20.08 below the horizontal (Figure 3-37). If the whale continues in a straight line for 150...

Question

13. A whale comes to the surface to breathe and then dives atan angle of 20.08 below the horizontal (Figure 3-37). If the whale continues in a straight line for 150 m, (a) how deep is it, and (b) how far has it traveled horizontally?20.09FIGURE 3-37 Problem 13

13. A whale comes to the surface to breathe and then dives atan angle of 20.08 below the horizontal (Figure 3-37). If the whale continues in a straight line for 150 m, (a) how deep is it, and (b) how far has it traveled horizontally? 20.09 FIGURE 3-37 Problem 13



Answers

A whale comes to the surface to breathe and then dives at an angle of $20.0^{\circ}$ below the horizontal (Figure $3-35$ ). If the whale continues in a straight line for $150 \mathrm{m},$ (a) how deep is it, and (b) how far has it traveled horizontally?

A whale takes a dive at an angle of 20 degrees below the horizontal and continues in a straight line for 215 meters. Here I've drawn a diagram of the whale diving 20 degrees below the horizontal and the high pot. News is the 215 meters that it continues in a straight line. The problem. Ask us, how deep does the whale dive and how far has a traveled horizontally? So for how far it has, how deep it has died, we won't have this part. Why and for how far has traveled horizontally will label as X now recall from trigonometry for why we can use yeah sine function, which gives us why over hi partners here we have the degree we have the angle which is 20 degrees, and the hypothesis which is 215 meters. So to find why we just multiply both sides by high pot news to get H signed data for why we can go ahead and plug in the numbers when we get a depth of about 51 meters Now for X, we do something similar, except we use co sign instead where co sign is X over high pot news here we solve for X plug in the numbers and arrive at an answer of 140 meters horizontally.

So we have a submarine going down and it made The depth down. They went down about 300 m and I can't draw that a little better. Now the ankle it made with the surface of the ocean was 21°.. So first one, I don't know how deep did they go? And that means I'm trying to find X. So if I put an X. There, How deep is it going to be? Well, I have 21° opposite and high partners. So I want the sign of 21° equals opposite over hypotenuse. So if I cross multiply I want 300 Times The Sine of 21. So my path is 107.5. So I can now put that in as 107.5 m. So we answered that one. How long is this horizontal path? So we now want to know that We can do it a couple ways. I can do. I have 21°.. I could do opposite An adjacent or I could do adjacent over hypotenuse. So why don't we do some things? We haven't done a co sign to the coastline of 21° equals adjacent Over hypotenuse. So all I need to do for that one is cross multiply 300 times the coastline of 21, 300 times The co sign of 21 equals 280 and we'll keep it at 280. So we now know that horizontal distance is 280 uh meters. And last but not least. What if they want to go to a depth of 1000 ft. So let's draw this. And now what we know is this is 1000 ft. Now remember no matter how deep they go, Hopes it's 1000 that still stays 280. So I want to know, how far do they go? And actually I better read the book because I can't remember the question, how far do they have to go? How far must go along its downward path? Okay, so the downward path is going to be this right here. We'll call that zee, It's this line right here. So well we know to 80. We know 1000. We can actually do the pythagorean theorem. 280 squared plus 1000 squared equals Z squared. So 280 Squared Plus 1000 squared equals. And let's take the square root of that. And we're going to get 1,038.5. And those are all of our answers.

And this problem. We have a bail shark that is swimming, and we know that they swim forward whenever they're ascending or descending and they move at a shallow angle. So this problem for part A, we want to know what was the horizontal distance between the sharks starting and ending positions. What was the total distance the shark swam and how much time did this take? So let's just go ahead and sketch the situations. We have the shark starting here and we have it swimming at an angle to the surface and then we'll call this H because this is a tight underwater. Do you the distance that it swam mhm and X, and we know that the angle above the horizontal data, you know, data is 13 degrees mhm were also given age. Did the shirt started at 50 m underwater and that it swam at a velocity of 3.85 m per second. So what was the world's alcohol distance between the sharks starting and ending positions? So we're looking for that X value so we can find not using the relation Engine of theater is opposite over adjacent. If we plug in the values that we can get X, so that gives us over X. And then here we want to solve for X so X is equal to H liberty and of data. We just plug in the values that we have for agent data here, which gives us an X value mhm 216.58 m. So that's the distance that the shark swam mhm between the horizontal distance. So now we want the total distance that the shark swam. And the best way to do that is using the Pythagorean theorem. So that's a squared. Plus B squared equals C squared and we're going to sub in for our values. So it's D squared Article two x squared plus each squared in this case, and we have mhm d squared and we're just gonna plug in the values moving that squared over to be the square roots we can solve mhm plugging in the value of this home for X. Okay, and then are you given for each and then once we solve that whatever distance value of 222.27 m, it's the total distance the short swam and then in part C. We want to know how much time this motion took, so we know that speed it is distance over time. But here we want to solve the time. So we set so we can rearrange and have time is equal to distance over speed and we know that the total distance that the shark swam was the 222.27 m and we know that it was swimming at a speed of 0.85 m per second. I mean, it's canceling and that leaves us with a total time that it took 261.49 seconds.

Hello students In this problem, we have given Ah hail SAARC's seem forward While ascending or descending, they seem along a straight line path at the shallows angle as they moved from the surface toe deepwater or from the depths to the surface. In one recorded DYP, ah sark started 50 m below the surface and swim at 0.85 m per second along a path tipped at a 30 degree angle have of the horizontal until reaching the surface first, we will scarce this situation. Suppose this is the surface of the water and initially supposed stark is here in deep water which is 50 m below that surface. This distance is 50 m and we have given this angle heater which is 13 degrees. Okay, so it it starts from here and goes up to this point on this surface. Now we have to find this horizontal distance. Suppose this is X Okay, We know that if this angle is a called a theater, then we can say this angle is also equal to theater. You can write 10 taters they called toe perpendicular, divided by base perpendicular. It's 50 and bases x. So from here we can get the Velo off X, which is 50 divided by 10 3 ties 13 degree. After solving this, we can get the value off eggs, which is a cool do 216 point 58 me that this is the for gentle distance traveled by the shark. Now we have to find the distance traveled by the SAARC. So this is the distance covered by that SAARC. So this distance, if we see the strangle if we see the strangle, this distance X is 200 and 16.58 And this distance suppose this is why is a call to 50 m We have to find this distance and this angle is also given. We know that this is a right triangle. According to the data, we're still wrong. We can write The square is equal to X squared the Spy Square. From here we can write This is equal to on the road access square. That's why square now This X is equal to 216 point 58 square plus Why is 50 m After solving this, we can get the value off B. It is a quarto. 222 point 27 meted. This is the distance covered by the shark. Now, in next part we have to find the time interval for this motion. We know that time is equal toe distance divided by you. Speed. Yeah. Dispense divided Very speed. So distances 200 22 When? 27 m. And the speed is we have given 0.85 m per second. So the time is after solving this, we get the time for this motion is to 61.5 seconds. They hope you understood this problem. Thank you.


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