Answer:
Incomplete question
Complete question: Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.
a. Determine the pre-collision momentum of the ball.
b. Determine the post-collision momentum of the ball.
c. Determine the momentum change of the ball.
Answer:
A. 1.5353kgm/s
B. 1.6963kgm/s
C. 0.161kgm/s
Step-by-step explanation:
A. The pre-collision momentum of the ball = mass of ball × velocity of ball
Mass of ball = 57.5g = 0.0575kg
Velocity of ball = 26.7m/s
Pre-collision momentum of ball = 0.0575×26.7
= 1.5353kgm/s
B. Post collision momentum of the ball = mass of ball × velocity of ball after impact
Velocity of ball after impact = 29.5m/s
Post collision momentum of ball after impact = 0.0575×29.5
= 1.6963kgm/s
C. Momentum change of ball = momentum after impact - momentum before imlact
= 1.6963kgm/s - 1.5353kgm/s
= 0.161kgm/s
Answer:
Step-by-step explanation:
is the given.
The instructions is to write as a single logarithm.
Let's use the power rule first:
Now we see the minus between the natural logs so we can use the quotient rule:
Assuming the distribution is continuous, you have
If instead the distribution is discrete, the value will depend on how the interval of number between 1 and 29 are chosen - are they integers? evenly spaced rationals? etc
Answer:
Step-by-step explanation:
(cos A+ cos B)-cos C
![=2cos \frac{A+B}{2}cos \frac{A-B}{2}-cos C~~~...(1)\\A+B+C=180\\A+B=180-C\\\frac{A+B}{2}=90-\frac{C}{2}\\cos \frac{A+B}{2}=cos(90-\frac{C}{2})=sin \frac{C}{2}\\cos C=1-2sin^2\frac{C}{2}\\(1)=2 sin \frac{C}{2} cos \frac{A-B}{2}-1+2sin^2\frac{C}2}\\=2sin\frac{C}{2}[cos \frac{A-B}{2}+sin \frac{C}{2}]-1~~~...(2)\\\\now~again~A+B+C=180\\C=180-(A+B)\\sin\frac{C}{2}=sin(90-\frac{A+B}{2})=cos \frac{A+B}{2}\\(2)=2sin\frac {C}{2}[cos \frac{A-B}{2}+cos \frac{A+B}{2}]-1\\](https://tex.z-dn.net/?f=%3D2cos%20%5Cfrac%7BA%2BB%7D%7B2%7Dcos%20%5Cfrac%7BA-B%7D%7B2%7D-cos%20C~~~...%281%29%5C%5CA%2BB%2BC%3D180%5C%5CA%2BB%3D180-C%5C%5C%5Cfrac%7BA%2BB%7D%7B2%7D%3D90-%5Cfrac%7BC%7D%7B2%7D%5C%5Ccos%20%5Cfrac%7BA%2BB%7D%7B2%7D%3Dcos%2890-%5Cfrac%7BC%7D%7B2%7D%29%3Dsin%20%5Cfrac%7BC%7D%7B2%7D%5C%5Ccos%20C%3D1-2sin%5E2%5Cfrac%7BC%7D%7B2%7D%5C%5C%281%29%3D2%20sin%20%5Cfrac%7BC%7D%7B2%7D%20cos%20%5Cfrac%7BA-B%7D%7B2%7D-1%2B2sin%5E2%5Cfrac%7BC%7D2%7D%5C%5C%3D2sin%5Cfrac%7BC%7D%7B2%7D%5Bcos%20%5Cfrac%7BA-B%7D%7B2%7D%2Bsin%20%5Cfrac%7BC%7D%7B2%7D%5D-1~~~...%282%29%5C%5C%5C%5Cnow~again~A%2BB%2BC%3D180%5C%5CC%3D180-%28A%2BB%29%5C%5Csin%5Cfrac%7BC%7D%7B2%7D%3Dsin%2890-%5Cfrac%7BA%2BB%7D%7B2%7D%29%3Dcos%20%5Cfrac%7BA%2BB%7D%7B2%7D%5C%5C%282%29%3D2sin%5Cfrac%20%7BC%7D%7B2%7D%5Bcos%20%5Cfrac%7BA-B%7D%7B2%7D%2Bcos%20%5Cfrac%7BA%2BB%7D%7B2%7D%5D-1%5C%5C)
Complimentary angles equal 90 degrees. So the answer would be C because 77+13= 90.
