Speed = (distance traveled) / (time to travel the distance).
Strange as it may seem, 'velocity' is completely different.
Velocity doesn't involve the total distance traveled at all.
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there. So the displacement in driving once around
any closed path is zero, because you end up where you started.
Velocity =
(displacement during some time)
divided by
(time for the displacement)
AND the direction from the start-point to the end-point.
For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):
Speed = (15km + 15km) / (10min + 5min) = (30/15) (km/min)
= 2 km/min.
Velocity = (end location - start position) / (15 min) = Zero .
Verrrrry interesting !
If the moon were replaced by something with a vastly greater mass
but at the same distance, then ...
-- The period of its revolution around the Earth would be much shorter.
That is, it would orbit the Earth in much less than 27.3 days. We might
see it go through a complete set of phases in 2 weeks, or even 1 week.
-- The ocean tides would be much greater. Low tides would be
much lower, and high tides would be much higher.
-- Sadly, the land tides, and the forces on the Earth's internal structure,
would also be much greater. That means great increases in earthquake
and volcanic activity.
-- The Earth and moon both revolve around their common center of
mass. Under the current arrangement ... with the Earth having 80 times
the mass of the Moon ... that point is inside the Earth, and it looks a lot
like the Moon is orbiting a stationary Earth.
When the new body arrives to replace the lightweight Moon, that point
will be a lot closer to the new companion ... maybe even inside it.
Then, it will look a lot like the monster is the stationary one, and the
Earth is orbiting it.
I actually don't believe that we would SEE that change, or feel it.
Answer: C
Explanation:
A collision in which the objects stick together is sometimes called “perfectly inelastic“
Newton’s Third Law describes two different bodies acting on each other.
now you've considered the action and reactions based on d ball, but have u considered the force the athlete exacts on the ground which is external to the athlete-ball system, now the ground in which the athlete stands exert a force back on the athelete-ball system which causes the ball to accelerate
Answer:

Explanation:
It is given that,
Charge of alpha particle, 
Mass of the alpha particle, 
Potential difference, 
Magnetic field, B = 2.5 T
The α-particle moves perpendicular to the magnetic field at all times. The initial speed of the alpha particle is 0 as it is at rest. Using the conservation of energy as :

is the speed of the α–particle




So, the speed of alpha particle is
. Hence, this is the required solution.