Answer:
2.64 m/s
Explanation:
Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna
Momentum = MV
Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s
Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s
Total momentum before collision = 1800 + 50 = 1850 kgm/s
Let's assume that they move together after collision. Then,
1850 = ( 600 + 100 ) V
1850 = 700V
V = 1850 / 700
V = 2.64285 m/s
Therefore, the momentum of the shark after collision is 2.64 m/ s approximately
<span>1. It must be an object which independently orbits the Sun (this means moons can't be considered planets, since they orbit planets)
2. It must have enough mass that its own gravity pulls it into a spheroidal shape.
3. </span><span>It must be large enough to "dominate" its orbit (i.e. its mass must be much larger than anything else which crosses its orbit).</span>
Vector it always consists to size(magnitude) and direction
Answer:
Explanation:
From, the given information: we are not given any value for the mass, the proportionality constant and the distance
Assuming that:
the mass = 5 kg and the proportionality constant = 50 kg
the distance of the mass above the ground x(t) = 1000 m
Let's recall that:
![v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}](https://tex.z-dn.net/?f=v%28t%29%20%3D%20%5Cdfrac%7Bmg%7D%7Bb%7D%2B%20%28v_o%20-%20%5Cdfrac%7Bmg%7D%7Bb%7D%29%5Ee%5E%7B-bt%2Fm%7D)
Similarly, The equation of mption:
![x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cdfrac%7Bmg%7D%7Bb%7Dt%2B%5Cdfrac%7Bm%7D%7Bb%7D%20%28v_o%20-%20%5Cdfrac%7Bmg%7D%7Bb%7D%29%20%281-e%5E%7B-bt%2Fm%7D%29)
replacing our assumed values:
where ![v_=0 \ and \ g= 9.81](https://tex.z-dn.net/?f=v_%3D0%20%5C%20and%20%5C%20g%3D%209.81)
![x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cdfrac%7B5%20%5Ctimes%209.81%7D%7B50%7Dt%2B%5Cdfrac%7B5%7D%7B50%7D%20%280%20-%20%5Cdfrac%7B%285%29%289.81%29%7D%7B50%7D%29%20%281-e%5E%7B-%2850%29t%2F5%7D%29)
![x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m](https://tex.z-dn.net/?f=x%28t%29%20%3D%200.981t%2B0.1%20%280%20-%200.981%29%20%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m)
![\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%28t%29%20%3D%200.981t-0.981%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m%7D)
So, when the object hits the ground when x(t) = 1000
Then from above derived equation:
![\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%28t%29%20%3D%200.981t-0.981%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m%7D)
![1000= 0.981t-0.981(1-e^{-(10)t}) \ m](https://tex.z-dn.net/?f=1000%3D%200.981t-0.981%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m)
By diregarding ![e^{-(10)t} \ m](https://tex.z-dn.net/?f=e%5E%7B-%2810%29t%7D%20%5C%20m)
![1000= 0.981t-0.981](https://tex.z-dn.net/?f=1000%3D%200.981t-0.981)
1000 + 0.981 = 0.981 t
1000.981 = 0.981 t
t = 1000.981/0.981
t = 1020.36 sec