let the vector Q is given as
![\vec Q = a\hat i + b\hat j + c\hat k](https://tex.z-dn.net/?f=%5Cvec%20Q%20%3D%20a%5Chat%20i%20%2B%20b%5Chat%20j%20%2B%20c%5Chat%20k)
given that
![P X Q = -6\hat j](https://tex.z-dn.net/?f=P%20X%20Q%20%3D%20-6%5Chat%20j)
here we know that
![P = 4\hat i + 3 \hatk](https://tex.z-dn.net/?f=P%20%3D%204%5Chat%20i%20%2B%203%20%5Chatk)
now by above equation
![(4\hat i + 3\hat k) X (a\hat i + b\hat j + c\hat k) = - 6\hat j](https://tex.z-dn.net/?f=%284%5Chat%20i%20%2B%203%5Chat%20k%29%20X%20%28a%5Chat%20i%20%2B%20b%5Chat%20j%20%2B%20c%5Chat%20k%29%20%3D%20-%206%5Chat%20j)
![4b\hat k - 4c\hat j + 3a\hat j - 3b\hat i = - 6\hat j](https://tex.z-dn.net/?f=4b%5Chat%20k%20-%204c%5Chat%20j%20%2B%203a%5Chat%20j%20-%203b%5Chat%20i%20%3D%20-%206%5Chat%20j)
so by comparing both sides
b = 0
4c - 3a = 6
also we know that
![a^2 + b^2 + c^2 = 17^2](https://tex.z-dn.net/?f=a%5E2%20%2B%20b%5E2%20%2B%20c%5E2%20%3D%2017%5E2)
![a^2 + 0 + (1.5 + 0.75a)^2 = 289](https://tex.z-dn.net/?f=a%5E2%20%2B%200%20%2B%20%281.5%20%2B%200.75a%29%5E2%20%3D%20289)
by solving above equation
a = 12.85 and c = 11.14
so the vector Q is given as
![Q = 12.85\hat i + 11.14\hat k](https://tex.z-dn.net/?f=Q%20%3D%2012.85%5Chat%20i%20%2B%2011.14%5Chat%20k)
Answer:
175 m
Explanation:
In a velocity vs time graph, displacement is the area under the curve.
We can calculate this as area of a trapezoid:
A = ½ (10 m/s + 60 m/s) (5 s)
A = 175 m
Or, we can split the area into a rectangle and a triangle.
A = (10 m/s) (5 s) + ½ (60 m/s − 10 m/s) (5 s)
A = 175 m
Answer:
it will get slower and eventually she will stop jumping because there isnt enough force on the gravity causing her to go up and down
Explanation:
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