Hi Professor Eman I hope all is well!! I wanted to let you know your videos have helped me out tremendously in biology last semester 🫶🏻 I was wondering do you have any resources that you recommend to use when studying for anatomy?
Wow that’s amazing! Thank you so much for watching my videos!! I don’t know much about different TH-cam channels that teach anatomy besides khan academy! I’m pretty sure there are many more but I’m not familiar with them
Hi Professor Eman, I may have gotten a bit confused but can you explain why for the t-total at minute 34:58 you multiplied the 3 seconds by 2? I may have missed it, and have been trying to understand why. Why would the answer not simply be 3 seconds?
3s is just the time to get to the maximum height (half the trip for the projectile). Then it takes another 3 seconds to fall back down. Since the question asks for the total projectile time, it’s 3*2=6s
Hello Professor Eman! I have been following your videos along with the Kaplan book. For the problem at 16:52, your final answer is 30.4 m, but the Kaplan book gives a final answer of 0.4 m because it didn't include the initial position (Yo) in the equation. Instead, it only used the equation: ΔX = VoT + (a(t²)/2). I’m a bit confused about which approach to follow since they give different answers. From my understanding, 0.4 m represents how much it went up, and 30 + 0.4 = 30.4 m would be the total. Essentially, it seems like one solution provides the total height, while the other focuses only on the displacement. Could you clarify this for me?
Hi Professor Eman, why did you replace mbg into the tension equation of block A ? Since block B does not have any friction in it? Thank you. around timestamp 57:00
@@professoreman2289 ok thank you I see.. I was wondering how come we couldn’t just use t=mbg but I see that the question stem spoke about static friction of block a..
as a visual learner studying for the mcat, these videos have helped me tremendously. thank you!
Omg that makes me very happy to hear!! Thank you for your kind comment ❤️❤️
Hi! Just want to say thank you so much for these videos, they are extremely helpful!
You’re so welcome queen! Glad to have you here❤️🥹
Hi Professor Eman I hope all is well!!
I wanted to let you know your videos have helped me out tremendously in biology last semester 🫶🏻
I was wondering do you have any resources that you recommend to use when studying for anatomy?
Wow that’s amazing! Thank you so much for watching my videos!! I don’t know much about different TH-cam channels that teach anatomy besides khan academy! I’m pretty sure there are many more but I’m not familiar with them
Hi Professor Eman, I may have gotten a bit confused but can you explain why for the t-total at minute 34:58 you multiplied the 3 seconds by 2? I may have missed it, and have been trying to understand why. Why would the answer not simply be 3 seconds?
3s is just the time to get to the maximum height (half the trip for the projectile). Then it takes another 3 seconds to fall back down. Since the question asks for the total projectile time, it’s 3*2=6s
@@professoreman2289 ahh that makes so much sense! thank you so much ! ◡̈
Of course! Let me know if you have any other questions:)
HI professor Eman, Where did you get hte -9.8 m/s^2 from in 11:22. JZK for everything !
Acceleration due to gravity
Hello Professor Eman! I have been following your videos along with the Kaplan book. For the problem at 16:52, your final answer is 30.4 m, but the Kaplan book gives a final answer of 0.4 m because it didn't include the initial position (Yo) in the equation. Instead, it only used the equation: ΔX = VoT + (a(t²)/2).
I’m a bit confused about which approach to follow since they give different answers. From my understanding, 0.4 m represents how much it went up, and 30 + 0.4 = 30.4 m would be the total. Essentially, it seems like one solution provides the total height, while the other focuses only on the displacement.
Could you clarify this for me?
They’re both the same answer technically. So 30.4 is the distance the ball is from the ground, while 0.4 is the reference from the ledge.
Hi Professor Eman, why did you replace mbg into the tension equation of block A ? Since block B does not have any friction in it? Thank you. around timestamp 57:00
As stated, T=fs and for block B which is staying still that means that T=mbg (this is true because the block is not moving) so therefore T=mbg=fs
@@professoreman2289 ok thank you I see.. I was wondering how come we couldn’t just use t=mbg but I see that the question stem spoke about static friction of block a..
Yup! Let me know if you have any follow up questions :)
can you do practice problems? thank you!
There is a practice problem video associated with ever chapter, look at the playlist :)