The asteroid's mass is so small that it has a much smaller acceleration
due to gravity than Earth has. That means that things weigh very very little
on the surface of an asteroid. It also means that the "escape velocity" from
an asteroid is very low, and orbital velocities are very low at any distance off
of its surface.
As an extreme example: You know how when you walk, you naturally rise up
on the toes of one foot while you reach out with the other one to take a step ?
All of those motions are what you learn in Earth's gravity. On an asteroid, that
natural action of rising up on your toes might launch you into a long, high arc,
like a golf ball. Or it might even exceed escape velocity and you'd sail up off
of the asteroid and never come back down to it.
The answer is C. The bacteria create identical replications of it former self.
Answer
given,
mass of bowling ball = 7.25 Kg
moving speed of the bowling ball = 9.85 m/s
mass of bowling in = 0.875 Kg
scattered at an angle = θ = 21.5°
speed after the collision = 10.5 m/s
angle of the bowling ball
![tan \theta_1 = \dfrac{-[m_2v_2Sin \theta_2]}{m_1v_1 - (m_2v_2cos \theta_2)}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5Bm_2v_2Sin%20%5Ctheta_2%5D%7D%7Bm_1v_1%20-%20%28m_2v_2cos%20%5Ctheta_2%29%7D)
![tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5B0.875%5Ctimes%2010.5%20%5Ctimes%20Sin%2021.5%5E0%5D%7D%7B7.25%5Ctimes%209.85%20-%20%280.875%5Ctimes%2010.5%20%5Ctimes%20cos%2021.5%5E0%29%7D)
![tan \theta_1 = \dfrac{-[3.3672]}{62.86}](https://tex.z-dn.net/?f=tan%20%5Ctheta_1%20%3D%20%5Cdfrac%7B-%5B3.3672%5D%7D%7B62.86%7D)
b) magnitude of final velocity
v = 8.68 m/s
Answer:
This is an example of Inelastic colission
Explanation:
Step one:
given:
mass of moose m1 = 620 kg
mass of train m2= 10,000kg
Initial velocity of moose u1= 0 m/s
Initial velocity of train v1 = 10m/s
combined velocity of the system is given as v
Applying the conservation of momentum equation we have
m1u1+ m2u1= (m1+m2)V
substitutting we have
620*0+10000*10= (620+10000)V
100000= 10620V
divide both sides by 10620
V = 100000/10620
V=9.41m/s
The velocity of the moose after impact is 9.41m/s
Sound travels 1,480 meters per second, which is about 4.3 times as fast as air. Sound travels much slowly in air. This has to do with the frequency, intensity and amplitude of waves, which are affected differently in water and air.
