With the addition of vectors we can find that the correct answer is:
C) Q> P > R = S > T
The addition of vectors must be done taking into account that they have modulus and direction. The analytical method is one of the easiest methods, the method to do it is:
- Set a Cartesian coordinate system
- Decompose vectors into their components in a Cartesian system
- Perform the algebraic sums on each axis
- Find the resultant vector using the Pythagoras' Theorem to find the modulus and trigonometry to find the direction.
In this exercise indicate that the modulus of all vectors is the same, suppose that the value of the modulus is A.
We fix a Cartesian coordinate system with the horizontal x axis and the vertical y axis, we can see that we do not need to perform any decomposition, so we perform the algebraic sums
Diagram P
x-axis
x = 2A
y-axis
y = 2A
The modulus of the resulting vector can be found with the Pythagorean Theorem
P = 
P = 
P = 2 √2 A
Diagram Q
x-axis
x = 3A
y-axis
y = A
Resulting
Q =
Q =
Q = 
Diagram R
x- axis
x = 0
y-axis
y = 2 A
Resulting
R =
R = 
Diagram S
x-axis
x = 2 A
y-axis
y = 0
Resulting
S = 2A
Diagram T
x- axis
x = 0
y-axis
y = 0
Resultant T = 0
We order the diagram from highest to lowest
Q> P> R = S> T
When reviewing the different answers, the correct one is:
C. Q> P> R = S> T
Learn more about adding vectors here:
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D=rt
when biker A catches biker B, the time they've been riding is the same, so
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters
Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
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I think that the answer is A
