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Why is dv/dt=dv/dx.v ?
I just want to see the steps, explained carefully.
Step by step
Solved in 2 steps with 2 images
- length period Vlength 2. Turn your intercept into a statement. Hint: "When the is 0, the is ." cm Vcm Vx BIU EE EE O O 1. 19 0.8625 4.358898943540674 24 0.9667 4.898979485566356 29 1.1083 5.385164807134504 34 1.1958 5.830951894845301 3. Does your intercept make sense? How do you know? BIU E=E E 39 1.2292 6.244997998398398 44 1.3625 6.6332495807108 49 1.4083 7 54 1.4542 7.3484692283495345 4. A represents the slope of your line of best fit. Pay attention to the number and the unit. Mass Curve: 58 1.5375 7.615773105863909 7.937253933193772 A: 1.378 LO 63 1.6792 8.246211251235321 RMSE : 4.029 g 11 68 1.6875 Take a look at the box below your graph to find the slope of your line. What is the slope of your line? period vs Vlength BIU E= EE 1.7 1.6 1.5 1.4 5. Turn your slope into a "for every" statement. Hint: The goes up for every 1 1.3 of ." BIU E= EE 1.2 1.1 1 0.9 6. Write out the equation for your line of best fit. Hint: Make sure you have: 4.5 5 5.5 6 6.5 7 7.5 8 • Variables that…Help, I believe the answer is B but I just want a second opinion on if I am correct.Answers that are incorrect are 96.537, 19.307, 193.07, 110.1, 48.260, 96.4, & 2413.0. Im not sure what I’m doing wrong, but any help is appreciated. Please solve part a of this problem that is attached below. Thank you!
- Ans:1395N Please show your complete solution and write your answer clearly. Thank you.Let A=(2i+6j-3k) m and B=(4i+2j+k) m . What is the product of A-B ? * (8i + 12j- 3k) m2 (12i- 14j- 20k) m² 23 m2 17 m 4.jpg 3.jpg A4What is wrong with the following expressions? How can you correct them? (a) C=AB , (b) C=AB , (c) C=AB , (d) C=AB , (e) C+2A=B , (f) C=AB , (g) AB=AB , (h) C=2AB , (i) C=A/B , and (j) C=A/B .
- 2v2 sin 0, cos 0; Kato asks why a 45° launch angle results in the maximum horizontal range. Abel uses R = to attempt an explanation. Which explanation is correct? |"I used the trigonometric identity sin(20) = 2 sin 0 cos 0 to rewrite the expression for the 2v, sin 0, cos 0,, and got R = V sin(20,). And, of course, when 0, = 45°, 20, = 90°, range, R = and sin 90° = 1, which is its maximum value." "Because sin 45° = cos 45° = 1, the maximum value for sine and cosine, R, is maximum when 0, = 45°." "I used the trigonometric identity, sin(20) = sin 0 cos 0, to rewrite the expression for the range, 2v? sin 0, cos , and got R = ŕ sin(20). When 0, = 45°, 20, = 90°, and sin 90° = 1, R = which is its maximum value." "This cannot be shown using an equation. The results are empirical, and we only know this from direct observation." Abel is correct. He knows that for angles greater than 45°, although the total time of flight of the projectile is longer, the horizontal component of velocity is less,…! Required information Given the function y(x, t) = (4.00 cm) sin ((378 rad/s)t - (314 rad/cm)x). Which of the following equations is correct? ✓ (Click to select) T = Vλ λ = ντ νελτPlease explain clearly. Thank you!
- Please answer letter B briefly and concisely po. Please make the illustration clear too. Thank you!?Hello, the text is a little fuzzy. Could you please provide a clearer version? Thank you so much. I will upvote immediately when the answer is readySolve the problem exactly like the example problem attached. The problem you're solving has a black background. Use equation Vf^2=Vi^2+2ax Givens (right train) X= Want t= ? Vi= Vf= a= -1 m/s Givens (left train) X= Want t= ? Vi= Vf= a= -0.5 m/s