A 3.0 X 10^3 grams rock swings in a circle with a diameter of 1000 cm. Given the constant speed around the circle is 17.90 mph (
1 answer:
Answer:
a = 12.8 m/s^2
Explanation:
To find the centripetal acceleration you use the following formula:
(1)
v: tangential speed of the rock = 17.90mph
r: radius of the orbit = 1000cm/2 = 500cm = 0.5m
You first change the units of the tangential velocity in order to replace the values of v and r in the equation (1):
Then, you use the equation (1):
hence, the centripetal acceleration is 12.8 m/s^2
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(a) Let's convert the final speed of the car in m/s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
Answer:
there yah go that's the answer
Answer:
m1/6 ( c )
Explanation:
since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy
After first collision
m1v = m1v1 + m2v2 --- ( 1 )
After second collision
m2v2 = m2v2 + m3v3 ---- ( 2 )
combining equations 1 and 2
m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )
All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3
note ; 3p = m1v ∴ m3 = ----- ( 4 )
applying conservation of energy
3v = v1 + v2 + v3 ------- ( 5 )
also 3m1v1 = m1v = v1 = v/3 =
v2 + v3 = 8/3 v ----- ( 6 )
next eliminate V3 for equation 6 by applying conservation of energy and momentum
m1 = 2m2 ------ ( 7 )
now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v
hence ; v2 = 2/3 v ------- ( 8 )
solving with equation 6 and 8
v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )
solving with equation 9 and 10
m3 = m1/3 * 1/2 = m1/6