Answer:
20.7
Explanation:
:0 because basis of the daily occured
Answer:
a) The speed of the slider is 4.28 in/s
b) The velocity vector is 2.33 in/s
Explanation:
a) According to the diagram 1 in the attached image:
Also:
![v_{C} =v_{A}+w_{AC}*r_{C/A}\\v_{Ci}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC} \\6.883&-9.829&0\end{array}\right]\\v_{Ci}=-3j+(0+9.829w_{AC} i-(0-6.883w_{AC})j\\v_{Ci}=9.829w_{AC}i+(-3+6.883w_{AC})j](https://tex.z-dn.net/?f=v_%7BC%7D%20%3Dv_%7BA%7D%2Bw_%7BAC%7D%2Ar_%7BC%2FA%7D%5C%5Cv_%7BCi%7D%3D-3j%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%260%26w_%7BAC%7D%20%5C%5C6.883%26-9.829%260%5Cend%7Barray%7D%5Cright%5D%5C%5Cv_%7BCi%7D%3D-3j%2B%280%2B9.829w_%7BAC%7D%20i-%280-6.883w_%7BAC%7D%29j%5C%5Cv_%7BCi%7D%3D9.829w_%7BAC%7Di%2B%28-3%2B6.883w_%7BAC%7D%29j)
If we comparing both sides of the expression:
b) According to the diagram 2 in the attached image:
![v_{C}=v_{A}+w_{AC}r_{C/A}\\v_{C}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC}\\7.713&-9.192&0\end{array}\right] \\v_{Ci}=-3j+(9.192w_{AC})i+7.713w_{AC}j\\v_{Ci}=9.192w_{AC}i+(7.713w_{AC}-3)j](https://tex.z-dn.net/?f=v_%7BC%7D%3Dv_%7BA%7D%2Bw_%7BAC%7Dr_%7BC%2FA%7D%5C%5Cv_%7BC%7D%3D-3j%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%260%26w_%7BAC%7D%5C%5C7.713%26-9.192%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5Cv_%7BCi%7D%3D-3j%2B%289.192w_%7BAC%7D%29i%2B7.713w_%7BAC%7Dj%5C%5Cv_%7BCi%7D%3D9.192w_%7BAC%7Di%2B%287.713w_%7BAC%7D-3%29j)
Comparing both sides of the expression:
![v_{B}=v_{C}+w_{AC}r_{B/C}\\v_{B}=3.57i+\left[\begin{array}{ccc}i&j&k\\0&0&0.388\\-3.856&4.59&0\end{array}\right] \\v_{B}=3.57i+(0-1.78)i-(0+1.499)j\\v_{B}=1.787i-1.499j\\|v_{B}|=\sqrt{1.787^{2}+1.499^{2} } =2.33in/s](https://tex.z-dn.net/?f=v_%7BB%7D%3Dv_%7BC%7D%2Bw_%7BAC%7Dr_%7BB%2FC%7D%5C%5Cv_%7BB%7D%3D3.57i%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%260%260.388%5C%5C-3.856%264.59%260%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5Cv_%7BB%7D%3D3.57i%2B%280-1.78%29i-%280%2B1.499%29j%5C%5Cv_%7BB%7D%3D1.787i-1.499j%5C%5C%7Cv_%7BB%7D%7C%3D%5Csqrt%7B1.787%5E%7B2%7D%2B1.499%5E%7B2%7D%20%20%7D%20%3D2.33in%2Fs)
The first one is at a higher freq
If a fire breaks out in the back of your boat you should put
the back of the boat into the wind so that the fire will not spread easily.
Then turn off the engine and paddle the boat to keep the boat into the wind.
Also use a fire extinguisher to put the fire off.
Answer:
a) Block 1 = 72.9kgm/s
Block 2 = 0kgm/s
b) vf = 1.31m/s
c) ∆KE = 936.36Joules
Explanation:
a) Momentum = mass× velocity
For block 1:
Momentum = 2.7×27
= 72.9kgm/s
For block 2:
Momentum = 53(0) (body is initially at rest)
= 0kgm/s
b) Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the block
u1 and u2 are their initial velocity
v is the common velocity
Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)
2.7(27)+53(0) = (2.7+53)v
72.9 = 55.7v
V = 72.9/55.7
Vf = 1.31m/s
c) kinetic energy = 1/2mv²
Kinetic energy of block 1 = 1/2×2.7(27)²
= 984.15Joules
Kinetic energy of block 2 before collision = 0kgm/s
Total KE before collision = 984.15Joules
Kinetic energy after collision = 1/2(2.7+53)1.31²
= 1/2×55.7×1.31²
= 47.79Joules
∆KE = 984.15-47.79
∆KE = 936.36Joules
