Answer:
angular speed of both the children will be same
Explanation:
Rate of revolution of the merry go round is given as
f = 4.04 rev/min
so here we have

here we know that angular frequency is given as



now this is the angular speed of the disc and this speed will remain same for all points lying on the disc
Angular speed do not depends on the distance from the center but it will be same for all positions of the disc
Answer:
2.25in³
Explanation:
For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³
See attached Table 314.16(B) from NEC 2011
Answer:
What does that even mean?
Explanation:
Answer:
1)ammeter
2)ised to check measure of current flow through a circuit
3)o.90 ambere
Answer:
The minimum speed required is 5.7395km/s.
Explanation:
To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

or

where
is the mass of the asteroid,
is its distance form earth's center,
is the mass of the earth, and
is the gravitational constant.
Solving for
we get:

putting in numerical values gives


in kilometers this is

Hence, the minimum speed required is 5.7395km/s.