For end of physics chapter 1 question #6 can you explain the equations used for each step and the steps needed to solve for this equation? I thought using the equations on page 15 figure 9 would help solve the problem but the answer explanation only describes using the main 4 kinematics equation.
Also which equation was used to help solve for time?
End of Chapter 1 Physics
Re: End of Chapter 1 Physics
Also for question # 7 I selected A because constant velocity means there is no acceleration but the correct answer was C and the explanation discussed net force but the question did not discuss force so I was confused on why force was related to the question. Can you further explain?

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Re: End of Chapter 1 Physics
#6:
The correct answer is D, and I believe the book states this too. You are correct in saying A, but both A and B are true, so answer choice D is correct. The acceleration of the car is constant and zero.
#7:
There are a bunch of ways to solve this, but the most straightforward is to just use vf^2 = vi^2 + 2ad, seeing as we have all the necessary ingredients for that (we have a final velocity, we have a distance and an acceleration  leaving us with only one unknown).
So, let's rearrange:
And that is it. Other expressions work too, and depending on which you use you might end up with a quadratic to solve for time. But this is certainly the recommended way.
The correct answer is D, and I believe the book states this too. You are correct in saying A, but both A and B are true, so answer choice D is correct. The acceleration of the car is constant and zero.
#7:
There are a bunch of ways to solve this, but the most straightforward is to just use vf^2 = vi^2 + 2ad, seeing as we have all the necessary ingredients for that (we have a final velocity, we have a distance and an acceleration  leaving us with only one unknown).
So, let's rearrange:
Code: Select all
vf^2  2ad = vi^2
(vf^22ad)^0.5 = vi
(20^24*75)^0.5 = vi
(400300)^0.5 = vi
10 = vi
Re: End of Chapter 1 Physics
Thank you, but I think you confused my questions lol.
But I meant #6 & #7 from the passage questions.
so #7
Patients moved through the dart throwing motions described in figure 2 with a constant rotational velocity. Given this, which of the following statements is true?
and #6
A patient is asked to throw a dart in order to hit the bullseye in the center of a target 5 m away. If the initial velocity of the dart is 10 m/s at 45° with the horizontal, from a height level with the bullseye, where will the dart hit the target?
But I meant #6 & #7 from the passage questions.
so #7
Patients moved through the dart throwing motions described in figure 2 with a constant rotational velocity. Given this, which of the following statements is true?
and #6
A patient is asked to throw a dart in order to hit the bullseye in the center of a target 5 m away. If the initial velocity of the dart is 10 m/s at 45° with the horizontal, from a height level with the bullseye, where will the dart hit the target?

 Posts: 616
 Joined: Sat Mar 30, 2019 8:39 pm
Re: End of Chapter 1 Physics
For #6:
viewtopic.php?f=12&t=2667&sid=69958a68d ... daea46bf91
For #7:
This is an oddly designed question and has a few flaws, but what it is trying to tell you is that if theta = 30 degrees, the horizontal component of the tension vector in the scapholunate will be half the total vector (by trigonometric properties: sin 30 = 0.5).
Therefore, for the horizontal forces to balance, the total tension in the scapholunate must be twice that in the radioscaphocapitate.
The big logical flaw here is of course that in a real system, the contact forces (normal reaction forces) between bones is really what keeps them stationary  tension on the ligaments does not necessarily have to all balance out. That would be impossibly unlikely, seeing as we can remain static in a variety of positions with very different angles between all our bones and ligaments.
viewtopic.php?f=12&t=2667&sid=69958a68d ... daea46bf91
For #7:
This is an oddly designed question and has a few flaws, but what it is trying to tell you is that if theta = 30 degrees, the horizontal component of the tension vector in the scapholunate will be half the total vector (by trigonometric properties: sin 30 = 0.5).
Therefore, for the horizontal forces to balance, the total tension in the scapholunate must be twice that in the radioscaphocapitate.
The big logical flaw here is of course that in a real system, the contact forces (normal reaction forces) between bones is really what keeps them stationary  tension on the ligaments does not necessarily have to all balance out. That would be impossibly unlikely, seeing as we can remain static in a variety of positions with very different angles between all our bones and ligaments.