I am like 98% sure it is C
Answer:
The recoil velocity of the projectile is 0.0025m/s
Explanation:
Given:
Mass of the projectile =0.005kg
Mass of the launcher = 1500kg
Velocity = 750 m/s.
To Find:
The recoil velocity of the projectile = ?
Solution:
The recoil velocity is the obtained by dividing the "recoil momentum" by the "mass of the recoil body". The recoil momentum is equal to the momentum of the other body.
The momentum of the other body is equal to it mass times its velocity.
Lets find the recoil momentum,
Recoil momentum = mass of the projectile X velocity
Recoil momentum =
Recoil momentum = 3.75
Now Recoil Velocity,
Recoil Velocity = 
Recoil Velocity = 
Recoil Velocity = 0.0025m/s
We could take the easy way out and just say
(110 kW) x (3 hours) = 330 kilowatt hours .
But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
________________________________________________
" 110 kw " means 110,000 watts = 110,000 joules/second .
(3 hours) x (3600 sec/hour) = 10,800 seconds.
(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules
That's
==> 1,188,000,000 joules
==> 1,188,000 kilojoules
==> 1,188 megajoules
==> 1.188 gigajoules
Atsa nawfulotta energy !
It goes back to that "110 kw appliance" that we started with.
That's no common ordinary household appliance. 110 kw is something like
147 horsepower. In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole. Most houses
are limited to 100 or 200 Amperes, tops. And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.
A. The fly's velocity to the ground is less than the car's velocity to the ground.
Answer:
A. Calculate vector
B. Update vector of each object
C. Update position of each object
Explanation:
Taking assumption of a system in which the forces are a function of the previous step's final position:
Firstly, we calculate the (vector) forces acting on the objects.
Secondly, Update the (vector) momentum of each object
(note: also update the velocity).
Thirdly, Update the (vector) position of each object.
The other operations are as follows;
i. select (dt),
ii. define mass,
iii. Put down constants,
iv. initialize variables, this would occur before the time-step loop is entered.
