It would be 20 protons left because 1-19= 20
Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J
Answer:
A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)
Explanation:
The moment, or impulse formula of the same forces acting on both car within 1 second is
In our case the forces are the same, the time duration of force acting on the cars are the same. Therefore, their momentum right after the force must also be the same.
B)
The speed of the cart changed because it stopped.
Hope I could help!
-Marshy