It depends on where the sun and earth is.
The answer would be:
D.
X: Low potential energy
Y: High Potential energy
Z: Flow of electrons
Z is clearly the flow of electrons, as shown by the arrow demonstrating the direction of the flow. So you can easily cross out choices B and C. Now, you can see that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.
Answer:
Part a)
Speed of the roller coaster is
Part b)
Since it is moving with non zero speed at some height above the ground
So we will have
Kinetic energy + Potential energy Both
Explanation:
As we know that there is no friction on the path
So here we can use mechanical energy conservation law
so we will have
Part a)
Part b)
Since it is moving with non zero speed at some height above the ground
So we will have
Kinetic energy + Potential energy Both
Answer:
V = 1.1 m/s
Explanation:
given,
mass of railroad car 1 , m = 30,000. kg
travelling at the speed , u = 2.2 m/s
mass of car 2, M = 30,000. kg
initial speed, u' = 0 m/s
final speed of the car after collision, V = ?
using conservation of momentum
m u + M u' = (M+m)V
30000 x 2.2 + 0 = (30000 + 30000) V
60000 V = 66000
V = 1.1 m/s
he velocity of the two cars is equal to V = 1.1 m/s
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
