Answer:
v = ![\left[\begin{array}{c}0.66&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0.66%260%5Cend%7Barray%7D%5Cright%5D)
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:
The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s
a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49
Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.
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This is incomplete question Complete Question is:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?
Answer:
A- Martin brings his friends home to meet grandpa.
Explanation:
took the test.
Answer:
There are so many questions which one you don't know
Answer:
A) 21.2 kg.m/s at 39.5 degrees from the x-axis
Explanation:
Mass of the smaller piece = 200g = 200/1000 = 0.2 kg
Mass of the bigger piece = 300g = 300/1000 = 0.3 kg
Velocity of the small piece = 82 m/s
Velocity of the bigger piece = 45 m/s
Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s
Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s
since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems
Resultant momentum² = 16.4² + 13.5² = 451.21
Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)
