Answer:
Her moment of inertia decreases causing her spin to speed up. The physics law behind this phenomenon is the conservation of angular momentum.
Explanation:
<em>Theory</em>
<u>The Law of conservation of angular momentum</u>
The angular momentum of a rotating body or a system remains constant unless it is acted upon by an external unbalance force.
Angular momentum = moment of inertia × angular velocity
The moment of inertia = mass×[perpendicular distance from axis of rotation]²
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⇒When skater draws in her outstretched arms the mass distribution get concentrated towards the axis of rotation so the moment of inertia of the body decrease.
But angular moment should conserve so angular velocity increases (spin increases)
Answer:
a) a = 19.0 m/s²
b) a = 2.9 m/s²
Explanation:
a) We draw the free body diagram of the box. There are 4 forces: the normal force N, the weight mg, the constant force F and the kinetic frictional force μ_kN. We can take the coordinate system which is rotated 55° from the horizontal, to ease the calculations. So, we write the equations of motion in each axis:
![x: F+mg\sin\theta-\mu_k N=ma\\y: N-mg\cos\theta=0 \implies N=mg\cos\theta\\](https://tex.z-dn.net/?f=x%3A%20F%2Bmg%5Csin%5Ctheta-%5Cmu_k%20N%3Dma%5C%5Cy%3A%20N-mg%5Ccos%5Ctheta%3D0%20%5Cimplies%20N%3Dmg%5Ccos%5Ctheta%5C%5C)
Substituting the expression for N in the first equation, we have:
![F+mg\sin\theta-\mu_kmg\cos\theta=ma\\\\\implies a=\frac{F+mg\sin\theta-\mu_kmg\cos\theta}{m}](https://tex.z-dn.net/?f=F%2Bmg%5Csin%5Ctheta-%5Cmu_kmg%5Ccos%5Ctheta%3Dma%5C%5C%5C%5C%5Cimplies%20a%3D%5Cfrac%7BF%2Bmg%5Csin%5Ctheta-%5Cmu_kmg%5Ccos%5Ctheta%7D%7Bm%7D)
If we plug in the given values, we have:
![a=\frac{190.0N+(15.0kg)(9.8m/s^{2})\sin55\°-0.300(15.0kg)(9.8m/s^{2})\cos55\° }{15.0kg} =19.0m/s^{2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B190.0N%2B%2815.0kg%29%289.8m%2Fs%5E%7B2%7D%29%5Csin55%5C%C2%B0-0.300%2815.0kg%29%289.8m%2Fs%5E%7B2%7D%29%5Ccos55%5C%C2%B0%20%7D%7B15.0kg%7D%20%3D19.0m%2Fs%5E%7B2%7D)
Since we chose the right-downward direction as positive, the positive sign in this case means that the box is accelerated downwards above the ramp.
b) In this case, the constant force F and the kinetic frictional force μ_kN point to the opposite side. In other words, we can just only change the sign of this two forces in the equations of part (a) and obtain:
![a=\frac{-F+mg\sin\theta+\mu_kmg\cos\theta}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B-F%2Bmg%5Csin%5Ctheta%2B%5Cmu_kmg%5Ccos%5Ctheta%7D%7Bm%7D)
Plugging in the given values:
![a=\frac{-190.0N+(15.0kg)(9.8m/s^{2})\sin55\°+0.300(15.0kg)(9.8m/s^{2})\cos55\° }{15.0kg} =-2.9m/s^{2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B-190.0N%2B%2815.0kg%29%289.8m%2Fs%5E%7B2%7D%29%5Csin55%5C%C2%B0%2B0.300%2815.0kg%29%289.8m%2Fs%5E%7B2%7D%29%5Ccos55%5C%C2%B0%20%7D%7B15.0kg%7D%20%3D-2.9m%2Fs%5E%7B2%7D)
Since we chose the right-downward direction as positive, the negative sign in this case means that the box is accelerated upwards above the ramp.
This means that the magnitude of the acceleration in this case is 2.9m/s².
Answer:
V2 = (V1 - u) / (1 - V1 u / c^2)
V1 = speed of ship in observer frame = .99 c to right
u = speed of frame 2 = -.99 c to left relative to observer
V2 = speed of V1 relative to V2
V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c
V2 = 1.98 / (1 + .99^2) c = .99995 c
Answer:
No, the ball will not clear the fence.
Solution:
Angular velocity, ![\omega = 70\ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2070%5C%20rad%2Fs)
Height, h = 1.2 m
Angle, ![\theta = 45^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2045%5E%7B%5Ccirc%7D)
Distance covered by the ball, d = 110 m
Length of the fence, l = 1.2 m
Radius of the axis, R = 46 cm = 0.46 m
Now,
To calculate the linear velocity of the ball, v:
![v = \omega R = 70\times 0.46 = 32.2\ m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20R%20%3D%2070%5Ctimes%200.46%20%3D%2032.2%5C%20m%2Fs)
Total time taken:
![t = \frac{2vsin\theta}{g} = \frac{2\times 32.2sin45^{\circ}}{9.8} = 4.646\ s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2vsin%5Ctheta%7D%7Bg%7D%20%3D%20%5Cfrac%7B2%5Ctimes%2032.2sin45%5E%7B%5Ccirc%7D%7D%7B9.8%7D%20%3D%204.646%5C%20s)
The distance at which the ball falls, with a = 0 is given by:
![x = vt + \frac{1}{2}at^{2} = 32.2cos45^{\circ}\times 4.646 = 105.78\ m](https://tex.z-dn.net/?f=x%20%3D%20vt%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D%20%3D%2032.2cos45%5E%7B%5Ccirc%7D%5Ctimes%204.646%20%3D%20105.78%5C%20m)
Since, the ball has to clear a fence 1.2 m long and a t a distance 110 m away, clearly it will not be able to cross it.