Answer:
For the first one, its "attract"
Part 1
If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position
given that
R = 1 m
g = 9.8 m/s^2
now from above equation we have
Part b)
for minimum value of angular speed we will have
The electrons float around in an outer sub shell
Answer:
v1 = 15.90 m/s
v2 = 8.46 m/s
mechanical energy before collision = 32.4 J
mechanical energy after collision = 32.433 J
Explanation:
given data
mass m = 0.2 kg
speed = 18 m/s
angle = 28°
to find out
final velocity and mechanical energy both before and after the collision
solution
we know that conservation of momentum remain same so in x direction
mv = mv1 cosθ + mv2cosθ
put here value
0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)
3.6 = 0.1765 V1 + 0.09389 v2 ................1
and
in y axis
mv = mv1 sinθ - mv2sinθ
0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)
0 = 0.09389 v1 - 0.1768 v2 .......................2
from equation 1 and 2
v1 = 15.90 m/s
v2 = 8.46 m/s
so
mechanical energy before collision = 1/2 mv1² + 1/2 mv2²
mechanical energy before collision = 1/2 (0.2)(18)² + 0
mechanical energy before collision = 32.4 J
and
mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²
mechanical energy after collision = 32.433 J
