Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
Answer:
Explanation:
The electric force on a charged particle is given by
where
q is the charge of the particle
E is the strength of the electric field
In this problem, we have
is the electric field strength
is the charge of the calcium ion
Therefore, the electric force exerted on the calcium ion is
Answer:
Explanation:
Using conservation law of momentum
let m₁ = mass of the railroad, initial u₁ = 3.49 m/s
let m₂ = mass of one of the coupled car, u₂= 1.28m/s
let m₃ = mass of the second car u₃ = 1.28 m/s
m₁u₁ + m₂u₂ + m₃u₃ = v ( m₁ + m₂ + m₃)
since the masses are the same
m₁ = m₂ = m₃
m ( 3.49 + 1.28 + 1.28) = 3m v
6.05 m = 3 mv
v = 6.05 m / 3m = 2.0167 m/s
b) kinetic energy lost = energy before collision - energy after collision
= (0.5m₁u₁² + 0.5 ( m₁+m₂) u₂² - 0.5 ( m₁ + m₂ + m₃) v
= (6.58365 + 1.7712) - 6.5951 = 1.76J
Answer:
C) If an ice cube is placed into a boiling water, then it will melt in less than 2 minutes.
Explanation:
Answer:
True........................
