Consider the vector b⃗ with magnitude 4.00 m at an angle 23.5∘ north of east. what is the x component bx of this vector? express
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Answer:
A) The free body diagrams for both the load of bricks and the counterweight are attached.
B) a = 2.96 m/s²
Explanation:
A)
The free body diagrams for both the load of bricks and the counterweight are attached.
B)
The acceleration of upward acceleration of the load of bricks is given by the following formula:
a = g(m₁ - m₂)/(m₁ + m₂)
where,
a = upward acceleration of load of bricks = ?
g = 9.8 m/s²
m₁ = heavier mass = mass of counterweight = 28 kg
m₂ = lighter mass = mass of load of bricks = 15 kg
Therefore, using these values in equation, we get:
a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)
<u>a = 2.96 m/s²</u>
Answer:(a)891.64 N
(b)0.7
Explanation:
Mass of crate
Crate slows down in
initial speed
inclination
From Work-Energy Principle
Work done by all the Forces is equal to change in Kinetic Energy
change in kinetic energy
(b)Coefficient of sliding friction
and
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
