Answer:
The acceleration of the boxes is 1.5 ft/s²
The displacement of the boxes during the speed-up period is 0.1875 ft.
Explanation:
Hi there!
Let´s convert the 45 ft/min into ft/s:
45 ft/min · 1 min/ 60 s = 0.75 ft/s
It takes the belt 0.5 s to reach this speed. Then, the acceleration of the boxes will be:
a = v/t
Where:
a = acceleration.
v = velocity.
t = time.
a = 0.75 ft/s / 0.5 s
a = 1.5 ft/s²
The acceleration of the boxes is 1.5 ft/s²
The equation of displacement is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the boxes at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since the origin of the frame of reference is located at the point where the boxes begin to move, x0 = 0. Since the boxes were initially at rest, v0 = 0. Then:
x = 1/2 · a · t²
x = 1/2 · 1.5 ft/s² · (0.5 s)²
x = 0. 1875 ft
The displacement of the boxes during the speed-up period is 0.1875 ft.
Answer:
Explanation:
Given
Mass of big box is M and small box is m
Tension T will cause the boxes to accelerate

where a=acceleration of the boxes
Now smaller box will slip over large box if the acceleration force will exceed the static friction
i.e. for limiting value


thus maximum tension

Answer:
m = 4.5021 kg
Explanation:
given,
Apparent mass of aluminium = 4.5 kg
density of air = 1.29 kg/m³
density of aluminium = 2.7 x 10⁷ kg/m³
true mass of the aluminium = ?
Weight in Vacuum
W = m g
W = ρV g
Air buoyancy acting on aluminium
B = ρ₀V g
Volume is the same in both cases since the volume of the aluminum
displaces an equal amount of volume air.
Apparent weight:
ρV g − ρ₀V g = 4.5 g
ρV − ρ₀V = 4.5

m = ρV


m = 4.5021 kg
Answer:
the answer is b
Explanation:
gravity pulls you down so on a scale you will weigh more. less gravity you will weigh less.