Answer:
A to C = 6.4 km
Explanation:
A to B = 4 km
B to C = 5 km
A to C = using pythagorean theorem
a² + b² = c²
a = A to B = 4
b = B to C = 5
c = A to C
c² = 4² + 5²
c = 6.4 km (A to C)
According to the scenario, the distance between town C from town A is found to be 6.40 Km.
Which background does this question depend on?The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.
According to the question,
The distance between town A to town B = 4 km.
The distance between town B to town C = 5 km.
Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:
[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].
[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]
[tex]AC^2[/tex] = 16 + 25 = 41.
AC = √41 = 6.40 km.
Therefore, the distance between town C from town A is found to be 6.40 Km.
To learn more about Pythagoras' theorem, refer to the link:
https://brainly.com/question/343682
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When you shift your focus, everything you
see is still in perfect focus.
True or false
Answer:
true
Explanation:
true
Answer:
I believe this is true
Explanation:
If your looking at something and you look at something else everything is still in perfect view and clear, in focus.
hope this helps :)
A wall 0.12 m thick having a thermal diffusivity of 1.5 × 10-6 m2/s is initially at a uniform temperature of 97°C. Suddenly one face is lowered to a temperature of 20°C, while the other face is perfectly insulated. Use the explicit finite-difference technique with space and time increments of 30 mm and 300 s to determine the temperature distribution at at 45 minutes.
Answer:
at t = 45 s :
To = 61.7⁰c, T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C
Explanation:
Wall thickness = 0.12 m
thermal diffusivity = 1.5 * 10^-6 m^2/s
Δt ( time increment ) = 300 s
Δ x = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )
using the explicit finite-difference technique
Detailed solution is attached below
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends and subject to a spatially varying distributed torsional load t(x) = p sin( 2π L x) , where p is a constant with units of torque per unit length. Determine the location and magnitude of the maximum internal torque in the bar.
Answer:
[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
Explanation:
Given that
Shear modulus= G
Sectional area = A
Torsional load,
[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]
For the maximum value of internal torque
[tex]\dfrac{dt(x)}{dx}=0[/tex]
Therefore
[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]
Thus the maximum internal torque will be at x= 0.25 L
[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]