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
What will happen if the sample is the
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When a sample of solid, liquid, or gas matter heats up, it expands. When matter gets hot, its particles gain kinetic energy. ... When matter cools down, its particles lose kinetic energy. The decreased kinetic energy lets the particles come closer together. The kinetic theory of matter can be used to explain how solids, liquids and gases are interchangeable as a result of increase or decrease in heat energy. ... If it is cooled the motion of the particles decreases as they lose energy.
Answer:
28,800m/p second
Explanation:
Calculate the distance per second so, 400m/50 s= 8m/p second now knowing the speed/hour and knowing an hour has 3,600 seconds,multiply it by 8 then you will get 28,800m/p second, or 28.8km/h
To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:

Where,
I = Moment of Inertia
= Angular Velocity
For the given object the moment of inertia is equivalent to

Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to




Our values are given as,

Replacing we have,


Therefore the angular speed after the catch slips is 0.2rad/s