The approximate lateral area of the prism is determined as 831 square inches.
<h3>
What is lateral area of the hexagonal prism?</h3>
The lateral area of the hexagonal prism is calculated as follows;
LA = PH
where;
- P is perimeter of the prism
- H is height
A = ¹/₂Pa
where;
- a is apothem = 10 inches
- A is base area = 346.41 in²
346.41 = ¹/₂(10)P
346.41 = 5P
P = 346.41/5
P = 69.282 inches
LA = PH
LA = 69.282 x 12
LA = 831.38 in²
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Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer:
Velocity
Explanation:
The slope of a position-time graph gives velocity of a moving object.
Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :
<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4
</span>
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Explanation:
the concept of conservation of the mechanical nerve
initial
Em₀ = 500 J
The energy is totally kinetic
Em₀ = K = ½ m v₀²
v₀ =
v₀ = √ (2 500/32)
v₀ = 5.59 m / s
v² = v₀² - 2 a x
the negative sign is because its stopping
a =
a = (5.59² - 5.1²) / 2 50
a = 0.0524 m / s²
Newton's second law
F = ma
F = 32 0.0524
F = 1.68 N
