Answer:
Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf
Explanation:
First we will find the acceleration of pickup truck.
As, the acceleration is uniform, therefore we can use Newton's third equation of motion:
2as = 
First convert speed into ft/sec
1 mile/hr = 1.47 ft/sec
therefore,
37 mile/hr = 37 x 1.47 ft/sec
37 mile/hr = 54.39 ft/sec
with initial speed 0 ft/sec (starting from rest), using in equation of motion:
a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)
a = 6.88 ft/sec²
Now, the total shear force will be given by Newton's second law of motion:
F = ma
F = (460 lbm +72 lbm)(6.88 ft/sec²)
F = 3660 lbf
Now for the max shear force in each of the two pegs we divide total fore by 2:
Force in each peg = F/2 = (3660 lbf)/2
<u>Force in each peg = 1830 lbf</u>
Answer:
Explanation:
According to Gauss's law, the electric flux through the circular plates is defined as the electric field multiplied by its area:
The magnetic field around the varying electric field of the circular plates is given by:
Replacing (1) in (2) and solving for 
:
In an exothermic reaction, there is a transfer of energy to the surroundings in the form of heat energy. The surroundings of the reaction will experience an increase in temperature. Many types of chemical reactions are exothermic, including combustion reactions, respiration & neutralization reactions of bases & acids.
Well heat comes out from the stove and moved to the pot and is TOUCHING. So the heat transfer would be conduction because the heat moves from one object to another by TOUCHING.
The similarity between them is they both involve heat, there is many but that's a simple one
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = 
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
