From the graph, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively
From the question, these are the given parameters;
Part A
The net work done will be the area under the graph.
From position X = 2m to X = 8m gives us the shape of a trapezium.
A = 1/2( a + b )h
A = 1/2( 2 + 6 ) x 8
A = 8 x 4
A = 32 Nm
From X = 0 to X = 2 gives us the shape of a triangle.
A = 1/2bh
A = 1/2 x 2 x (-4)
A = -4 x 1
A = -4 Nm
Net Work done = 32 - 4
Net work done = 28 Nm
Part B
Applying the concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m
Net Work done = 1/2m
Substitute all the necessary parameters
28 = 1/2 x 5.5 x 
5.5
= 56
= 56/5.5
= 10.18
V = 
V = 3.19 m/s
Therefore, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively
Learn more about work done here: brainly.com/question/8119756
Answer:
B) Electrons and protons
Explanation:
Each proton has a positive charge, and each electron has a negative charge. If the atom is electrically neutral, there must be the same number of electrons and protons.
A) and C) are wrong. Neutrons have no charge, so they can't counteract that of the protons or the electrons.
D) is wrong. The atomic number is the number of protons, and the mass number is the number of protons plus neutrons. The neutrons can't counteract the charge of the protons,
<span>When barium hydroxide is mixed with ammonium chloride, ice crystals form on the outside of the container.
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<span>Two or more atoms joined into a single particular form </span><span>a molecule</span>
Answer: 45.3°
Explanation:
Given,
Length of ladder = l
Weight of ladder = w
Coefficient of friction = μs = 0.495
Smallest angle the ladder makes = θ
If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,
N1 = mg and
N2 = μmg
Moment at a point A in the clockwise direction is
N2 Lsinθ - mg.(L/2).cosθ = 0
μmgLsinθ - mg.(L/2).cosθ = 0
μmgLsinθ = mg.(L/2).cosθ
μsinθ = cosθ/2
sin θ / cos θ = 1 / 2μ
Tan θ = 1 / 2μ
Substituting the value of μ = 0.495, we have
Tan θ = 1 / 2 * 0.495
Tan θ = 1 / 0.99
Tan θ = 1.01
θ = tan^-1(1.01)
θ = 45.3°