Answer:
Toshi must begin his walk at 11:00 AM in order that he can return by 8:00 PM.
Step-by-step explanation:
Since the Gotemba walking trail up Mount Fuji is about 9km long, and walkers need to return from the 18km walk by 8pm, if Toshi estimates that he can walk up the mountain at 1.5km / h on average, and down at twice that speed , these speeds taking into account meal breaks and rest times, to determine what is the latest time he can begin his walk so that he can return by 8pm the following calculation must be performed:
Climb: 1.5 km / h
Descent: 2 x 1.5 km / h = 3 km / h
Climb: 9 km / 1.5 km / h = 6 hours
Descent: 9km / 3 km / h = 3 hours
Total: 9 hours
8 PM = 20:00
20:00 - 09:00 = 11:00
Thus, Toshi must begin his walk at 11:00 AM in order that he can return by 8:00 PM.
Answer:
A) C1 = 0.00187 m = 0.187 cm, C2 = 0.0062 m = 0.62 cm
B) A sample of how the graph looks like is attached below ( periodic sine wave )
C) w =
is when the amplitude of the forced response is maximum
Step-by-step explanation:
Given data :
mass = 5kg
length of spring = 10 cm = 0.1 m
f(t) = 10sin(t) N
viscous force = 2 N
speed of mass = 4 cm/s = 0.04 m/s
initial velocity = 3 cm/s = 0.03 m/s
Formulating initial value problem
y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m
spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m
f(t) = 10sin(t/2) N
using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion
the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)
A) finding the solution of the initial value
attached below is the solution and
B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like
C attached below
Answer:
true
Step-by-step explanation:
Yes this is true ...vergetables give u more energy and strength but they also make your healthier and stronger.
(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)
Use the shell method. Each shell has a height of 3 - 3/4 <em>y</em> ², radius <em>y</em>, and thickness ∆<em>y</em>, thus contributing an area of 2<em>π</em> <em>y</em> (3 - 3/4 <em>y</em> ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ <em>y</em> ≤ 2, thus given by the integral

Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 <em>x</em>), thus contributing an area of <em>π</em> (√(4/3 <em>x</em>))² = 4<em>π</em>/3 <em>x</em>. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ <em>x</em> ≤ 3, or by the integral

Using either method, the volume is 6<em>π</em> ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...