<h3>Hi there !</h3><h2>Option A is correct </h2><h3> Please refer the attachment for explanation</h3><h2>Stay safe, stay healthy and blessed</h2><h2>Have a marvelous day</h2><h2>Thank you</h2>
This chart shows characteristics of three different types of waves.
1 answer:
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To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load = 10000 kg
mass of flat bed = 20000 kg
initial speed of truck = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs -------------let this be equation 1
where = normal force = mg
so
Fs,max = μs mg
ma = μs mg
divide through by mass
a = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a = 0.5 × 9.8 m/s²
a = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
² =
² + 2aΔx
where is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load