By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and
1 answer:
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Answer:
The found acceleration in terms of h and t is:
Explanation:
(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)
We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.
<h3></h3><h3>Stage 1</h3>Constant acceleration, starts from rest.
Distance =
Velocity =
Constant velocity where
Velocity =
Distance =
<h3>Constant deceleration where
Velocity =
Distance =
Total height = y₁ + y₂ + y₃
Total height =
Find acceleration by rearranging the found equation of total height.
Total Height = h
h = 5a(t₁)²
Answer:
a) 1.301 kg/s
b) 0.001301 m³/s
c) V₁ = 6.505 m/s, V₂ = 1.626 m/s
d) 118.93 kPa
Explanation:
Given:
The number of cans = 220
The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³
time = 1 minute = 60 seconds
gauge pressure at point 2, P₂ = 152 kPa
b) Thus, the volume flow rate, Q = Volume/ time
Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s
a) mass flow rate = Volume flow rate × density
since it is mostly water, thus density of the drink = 1000 kg/m³
thus,
mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s
c) Given:
Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²
Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²
also,
Q = Area × Velocity
thus, for point 1
0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)
or
V₁ = 6.505 m/s
for point 2
0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)
or
V₂ = 1.626 m/s
d) Applying the Bernoulli's theorem between the points 1 and 2 we have
or
on substituting the values in the above equation, we get
it is given that point 1 is above point 2 thus, y₂ -y₁ is negative
or
thus, gauge pressure at point 1 is 118.93 kPa
Answer:
the force P required for impending motion is 132.3 N
the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
Explanation:
Given that:
mass of the cabinet m = 45 kg
coefficient of static friction μ = 0.30
A free flow body diagram illustrating what the question represents is attached in the file below;
The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.
Thus; considering the forces along the vertical axis ; we have :
The upward force and the downward force is :
where;
are the normal contact force at center point A and B respectively .
------- equation (1)
Considering the forces on the horizontal axis:
where ;
are the static friction at center point A and B respectively.
which can be written also as:
replacing our value from equation (1)
P = 0.30 ( 441)
P = 132.3 N
Thus; the force P required for impending motion is 132.3 N
b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm
the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at and it is shifted to
:
Then:
h = 0.8 m
Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
