Answer:
The observable universe is still huge, but it has limits. because it's most likely like an plane all round.
Explanation:
Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:
![m_1v_1+m_2v_2=(m_1+m_2)v_f](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29v_f)
Our values are given by,
![m_1=m_2=72Kg](https://tex.z-dn.net/?f=m_1%3Dm_2%3D72Kg)
As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:
![v_{x1}=5.45m/s](https://tex.z-dn.net/?f=v_%7Bx1%7D%3D5.45m%2Fs)
![v_{y1}=0](https://tex.z-dn.net/?f=v_%7By1%7D%3D0)
Skate 2:
![v_{x2} = 5.45*cos105= -1.41m/s](https://tex.z-dn.net/?f=v_%7Bx2%7D%20%3D%205.45%2Acos105%3D%20-1.41m%2Fs)
![v_{y2} = 5.45*sin105 = 5.26m/s](https://tex.z-dn.net/?f=v_%7By2%7D%20%3D%205.45%2Asin105%20%3D%205.26m%2Fs)
Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:
![m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}](https://tex.z-dn.net/?f=m_1v_%7By1%7D%2Bm_2v_%7By2%7D%3D%28m_1%2Bm_2%29v_%7Bfy%7D)
![0+75*5.25=(75+75)v_{fy}](https://tex.z-dn.net/?f=0%2B75%2A5.25%3D%2875%2B75%29v_%7Bfy%7D)
![v_{fy}=\frac{5.25}{2}](https://tex.z-dn.net/?f=v_%7Bfy%7D%3D%5Cfrac%7B5.25%7D%7B2%7D)
![v_{fy}=2.63m/s](https://tex.z-dn.net/?f=v_%7Bfy%7D%3D2.63m%2Fs)
Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
The answer to your question would be team C, because the lifted the most weight in the shortest time. Team A might have been the fastest team but the also lifted the least amount of weight. And team B lifted a good amount of weight but they also did it the slowest.
Answer:
a. 12 m/s² down
Explanation:
Acceleration has units of length per time squared. Acceleration is a vector, so it also has a direction.
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, ![\alpha=3.3 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%3D3.3%20rad%2Fs%5E2)
Diameter of the wheel, d=21 cm
Radius of wheel,
cm
Radius of wheel, ![r=\frac{21\times 10^{-2}}{2} m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B21%5Ctimes%2010%5E%7B-2%7D%7D%7B2%7D%20m)
1m=100 cm
Magnitude of total linear acceleration, a=![1.7 m/s^2](https://tex.z-dn.net/?f=1.7%20m%2Fs%5E2)
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,![a_t=\alpha r](https://tex.z-dn.net/?f=a_t%3D%5Calpha%20r)
![a_t=3.3\times \frac{21\times 10^{-2}}{2}](https://tex.z-dn.net/?f=a_t%3D3.3%5Ctimes%20%5Cfrac%7B21%5Ctimes%2010%5E%7B-2%7D%7D%7B2%7D)
![a_t=34.65\times 10^{-2}m/s^2](https://tex.z-dn.net/?f=a_t%3D34.65%5Ctimes%2010%5E%7B-2%7Dm%2Fs%5E2)
Radial acceleration,![a_r=\frac{v^2}{r}](https://tex.z-dn.net/?f=a_r%3D%5Cfrac%7Bv%5E2%7D%7Br%7D)
We know that
![a=\sqrt{a^2_t+a^2_r}](https://tex.z-dn.net/?f=a%3D%5Csqrt%7Ba%5E2_t%2Ba%5E2_r%7D)
Using the formula
![1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}](https://tex.z-dn.net/?f=1.7%3D%5Csqrt%7B%2834.65%5Ctimes%2010%5E%7B-2%7D%29%5E2%2B%28%5Cfrac%7Bv%5E2%7D%7Br%7D%29%5E2%7D)
Squaring on both sides
we get
![2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}](https://tex.z-dn.net/?f=2.89%3D1200.6225%5Ctimes%2010%5E%7B-4%7D%2B%5Cfrac%7Bv%5E4%7D%7Br%5E2%7D)
![\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E4%7D%7Br%5E2%7D%3D2.89-1200.6225%5Ctimes%2010%5E%7B-4%7D)
![v^4=r^2\times 2.7699](https://tex.z-dn.net/?f=v%5E4%3Dr%5E2%5Ctimes%202.7699)
![v^4=(10.5\times 10^{-2})^2\times 2.7699](https://tex.z-dn.net/?f=v%5E4%3D%2810.5%5Ctimes%2010%5E%7B-2%7D%29%5E2%5Ctimes%202.7699)
![v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}](https://tex.z-dn.net/?f=v%3D%28%2810.5%5Ctimes%2010%5E%7B-2%7D%29%5E2%5Ctimes%202.7699%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D)
![v=0.418 m/s](https://tex.z-dn.net/?f=v%3D0.418%20m%2Fs)
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s