Pragmatism: An approach that assesses the truth of meaning of theories or beliefs in terms of the success of their practical application.
Religion: Philosophy of religion is the philosophical study of the meaning and nature of religion. It includes the analyses of religious concepts, beliefs, terms, arguments, and practices of religious adherents. The scope of much of the work done in philosophy of religion has been limited to the various theistic religions.
Answer:
is the density of the substance which has 5gm and in an area of 12cm, 10cm and 2cm.
<u>Explanation:</u>
To find the density of a substance, we know that density is the ratio of mass of the substance and volume of the substance. Hence,
given that mass of the substance is 5gm and volume of the substance is 12cm×10cm×2cm =240 ![\mathrm{cm}^{3} \text { . density }=\frac{s}{240}=0.0208 \mathrm{gm} / \mathrm{cm}^{3}](https://tex.z-dn.net/?f=%5Cmathrm%7Bcm%7D%5E%7B3%7D%20%5Ctext%20%7B%20.%20density%20%7D%3D%5Cfrac%7Bs%7D%7B240%7D%3D0.0208%20%5Cmathrm%7Bgm%7D%20%2F%20%5Cmathrm%7Bcm%7D%5E%7B3%7D)
Therefore, we came to know that density of the substance which has mass and volume can be found by using the above formula. <em>
</em>is the density of the substance whose <em>mass is 5gm and volume is 240
.</em>
Answer:
The current at t= 0 sec, is 0 A
The current at t= 0.5 sec, is 2.2 A
The current at t= 1 sec, is 4.4 A
Explanation:
Given that
q(t) = 2.2 t²
We know that:- the change in the charge w.r.t. time is known as current. So,
![I=\dfrac{dq}{dt}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bdq%7D%7Bdt%7D)
q(t) = 2.2 t²
![\dfrac{dq}{dt}= 4.4 t](https://tex.z-dn.net/?f=%5Cdfrac%7Bdq%7D%7Bdt%7D%3D%204.4%20t)
I= 4.4 t
1.
t = 0 s :
I = 4.4 x 0 = 0 A
<u>Therefore, the current at t= 0 sec, is 0 A</u>
2 .
t= 0.5 s :
I = 4.4 x 0.5 = 2.2 A
<u>Therefore, the current at t= 0.5 sec, is 2.2 A</u>
3.
t= 1 s
I = 4.4 x 1 =4.4 A
<u>Therefore, the current at t= 1 sec, is 4.4 A</u>
Answer:
You can adjust the stroke width as small as you like. Then expand the stroke. Apply Layer > Expand Stroke. It makes the curve a filled area which looks the same as the curve.
Explanation:
Answer:
q₀ = 350,740.2885 N/m
Explanation:
Given
![q(x)=\frac{x}{L} q_{0}](https://tex.z-dn.net/?f=q%28x%29%3D%5Cfrac%7Bx%7D%7BL%7D%20q_%7B0%7D)
σ = 120 MPa = 120*10⁶ Pa
![L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\](https://tex.z-dn.net/?f=L%3D4%20m%5C%5Cw%3D200%20mm%3D0.2m%5C%5Ch%3D300%20mm%3D0.3m%5C%5Cq_%7B0%7D%3D%3F%20%5C%5C)
We can see the pic shown in order to understand the question.
We apply
∑MB = 0 (Counterclockwise is the positive rotation direction)
⇒ - Av*L + (q₀*L/2)*(L/3) = 0
⇒ Av = q₀*L/6 (↑)
Then, we apply
![v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x](https://tex.z-dn.net/?f=v%28x%29%3D%5Cint%5Climits%5EL_0%20%7Bq%28x%29%7D%20%5C%2C%20dx%5C%5Cv%28x%29%3D-%5Cfrac%7Bq_%7B0%7D%7D%7B2L%7D%20x%5E%7B2%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7D%20%5C%5CM%28x%29%3D%5Cint%5Climits%5EL_0%20%7Bv%28x%29%7D%20%5C%2C%20dx%3D-%5Cfrac%7Bq_%7B0%7D%7D%7B6L%7D%20x%5E%7B3%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7Dx)
Then, we can get the maximum bending moment as follows
![M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m](https://tex.z-dn.net/?f=M%27%28x%29%3D0%5C%5C%20%28-%5Cfrac%7Bq_%7B0%7D%7D%7B6L%7D%20x%5E%7B3%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7Dx%29%27%3D0%5C%5C%20-%5Cfrac%7Bq_%7B0%7D%7D%7B2L%7D%20x%5E%7B2%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7D%3D0%5C%5Cx%5E%7B2%7D%20%3D%5Cfrac%7BL%5E%7B2%7D%7D%7B3%7D%5C%5C%20%20x%3D%5Csqrt%7B%5Cfrac%7BL%5E%7B2%7D%7D%7B3%7D%7D%20%3D%5Cfrac%7BL%7D%7B%5Csqrt%7B3%7D%20%7D%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7Dm)
then we get
![M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}](https://tex.z-dn.net/?f=M%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%3D-%5Cfrac%7Bq_%7B0%7D%7D%7B6%2A4%7D%20%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%5E%7B3%7D%2B%5Cfrac%7Bq_%7B0%7D%20%2A4%7D%7B6%7D%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%5C%5C%20M%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%3D-%5Cfrac%7B8%7D%7B9%5Csqrt%7B3%7D%20%7D%20q_%7B0%7D%20%2B%5Cfrac%7B8%7D%7B3%5Csqrt%7B3%7D%20%7D%20q_%7B0%7D%3D%5Cfrac%7B16%7D%7B9%5Csqrt%7B3%7D%20%7D%20q_%7B0%7Dm%5E%7B2%7D)
We get the inertia as follows
![I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bw%2Ah%5E%7B3%7D%20%7D%7B12%7D%20%5C%5C%20I%3D%5Cfrac%7B0.2m%2A%280.3m%29%5E%7B3%7D%20%7D%7B12%7D%3D4.5%2A10%5E%7B-4%7Dm%5E%7B4%7D)
We use the formula
σ = M*y/I
⇒ M = σ*I/y
where
![y=\frac{h}{2} =\frac{0.3m}{2}=0.15m](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bh%7D%7B2%7D%20%3D%5Cfrac%7B0.3m%7D%7B2%7D%3D0.15m)
If M = Mmax, we have
![(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}](https://tex.z-dn.net/?f=%28%5Cfrac%7B16%7D%7B9%5Csqrt%7B3%7D%20%7Dm%5E%7B2%7D%20%29%20q_%7B0%7D%5Cleq%20%5Cfrac%7B120%2A10%5E%7B6%7DPa%2A4.5%2A10%5E%7B-4%7Dm%5E%7B4%7D%20%20%20%7D%7B0.15m%7D%5C%5C%20q_%7B0%7D%5Cleq%20350%2C740.2885%5Cfrac%7BN%7D%7Bm%7D)