A man weighs 650 N. Inside a moving elevator, the floor exerts an upward force of 620 N on the man. What is the magnitude and di
2 answers:
Answer:Man's acceleration is 0.452 in downward direction
Explanation:
<em>Weight of man, W= 650 N</em>
<em>Normal force exerted by floor of elevator , N= 620 N</em>
<em>Since weight , W = mg</em>
<em>mass m= </em>
<em>Using Newtons 2nd law</em>
<em>W-N= ma where a= acceleration of man in the downward direction</em>
=> 650-620=
=> a= 0.452 <em>in downward direction </em>
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Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i =
where R is the resistance
R =
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i =
q =
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C