most of earth water a. contains salt b.is freshwater c. is frozen in glaciers d. is found underground
2 answers:
You might be interested in
Answer:
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
Explanation:
To solve the problem/ we first write the differential equation governing the motion. So,
with m = 1 slug and k = 9 lb/ft, the equation becomes
The characteristic equation is
D² + 9 = 0
D = ±√-9 = ±3i
The general solution of the above equation is thus
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6. So,
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
Answer:
a) 61.224 m
b) t = 7.070 seconds
c) horizontal component = 20 m/s; vertical component = 34.641 m/s
Note: I rounded all of these values to the nearest thousandth but if you want the precise values please read the explanation below.
Explanation:
<h2><u>Horizontal and Vertical Components:</u></h2>Let's start this problem by solving for the horizontal and vertical components of the initial velocity vector.
We can solve for these x- and y-components by using the formulas:
- h. component:
- v. component:
Where is the initial velocity (here it's given to us: 40 m/s) and
is the angle at which the object is launched above the horizontal (it's also given to us: 60°).
Substitute these given values into the formulas to solve for the horizontal and vertical components:
- h. component =
- v. component =
Input these values into a calculator and you will get:
- h. component = 20 m/s
- v. component = 34.641 m/s
Now we want to solve for the time t of the object before finding the maximum height of the object. In other words, the max height of the object is its vertical displacement at half of the time t we're about to find.
In order to solve for t, we can use one of the constant acceleration equations we are given in Physics. This equation is:
The time t is always solved for by using the vertical (y-direction) motion of the object in projectile motion, so therefore, we are going to be using this equation in terms of the y-direction.
Time is the same regardless of the x- or y- direction.
Now, we don't necessarily know the final velocity of the projectile, but we do know its final velocity in the y-direction at the very top of the trajectory, which is 0 m/s.
We can use this to our advantage and solve for only half of the time t, then multiply it by 2 at the end to get the full time that the object is in the air.
We have already solved for , which is the vertical component. We know that an object in projectile motion has an acceleration of -g in the y-direction, so we use -9.8 m/s² for a.
Subtract the vertical component from both sides of the equation.
Divide both sides of the equation by -9.8 in order to solve for t.
Remember that this is only half of the time that the object spends in the air. However, this is the time that it takes for the object to reach its maximum height, which we will use later. For now, let's say that the time of the object is 2t.
In order to find the maximum height of the object, let's use another kinematic equation for constant acceleration:
Since we are still dealing with the y-direction, we can change this equation to be in terms of y.
The displacement in the y-direction, or the vertical displacement, can be modeled by subtracting from both sides of the equation.
In order to solve for the maximum height of the object, we want to use the time t that it takes for the object to reach its highest point, which we found was ~3.53 seconds. This is true because the object essentially follows the movement of a parabola.
We know the vertical component , and we know the acceleration in the y-direction is -g, so let's substitute these values into the formula for vertical displacement and solve for
.
The maximum height of this object in projectile motion is 61.224 m.
(This answer exceeded the character limit if I included the "Helpful Shortcuts" section, so I included it as an attachment in case you're interested.)
