5. In which image below is the most work being wasted as heat?
1 answer:
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Answer:
- < 25 m/s
- triangle inequality
- between north and east
- 45° < angle < 60°
Explanation:
(a) Just as one-dimensional numbers add on a number line by putting them end-to-end, so two-dimensional numbers add on a coordinate plane the same way.
Here, we choose to let the positive y-axis represent North, and the positive x-axis, East. This is the way a map is conventionally oriented. The velocity of the plane is represented by a vector pointing north (up). Its length represents the magnitude of the velocity. Likewise, the wind is represented by a vector of length 15 pointing east (right). The sum of these is the hypotenuse of the triangle they form.
The magnitude of the sum can be found here using the Pythagorean theorem, but for the purpose of this question, you're not asked to find that.
Instead, you're asked to estimate whether it is more or less than 25 (m/s).
Your knowledge of the triangle inequality will tell you that the hypotenuse (resultant) must be shorter than the sum of the lengths of the sides of the triangle, hence must be less than 10+15 = 25.
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(b) The triangle inequality says the resultant is less than the sum of the other two sides of the triangle.
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(c) Since the wind is blowing the plane toward the east, but the plane is traveling toward the north, the resulting direction is somewhere between north and east.
__
(d) "Somewhere between north and east" can be expressed as the inequality ...
0° < angle < 90°
Answer:
V = 152.542 volts
Explanation:
Given data:
area of plates
distance between the plates is
charge =
we know that capacitance is given as
potential difference is given as
V = 152.542 volts
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
- ∑Fₓ = maₓ
- F_f - F_g sinΘ = maₓ
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
- T₁ = m₂(a + g)
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
