The mechanical energy of the car is ![1.54\cdot 10^6 J](https://tex.z-dn.net/?f=1.54%5Ccdot%2010%5E6%20J)
Explanation:
The mechanical energy of an object is equal to the sum of its gravitational potential energy (GPE) and its kinetic energy (KE):
E = GPE + KE
The GPE of the car is given by:
![GPE = mgh](https://tex.z-dn.net/?f=GPE%20%3D%20mgh)
where
m = 1000 kg is the mass of the car
is the acceleration of gravity
h = 30 m is the height of the car above the ground
Substituting,
![GPE=(1000)(9.8)(30)=2.94\cdot 10^5 J](https://tex.z-dn.net/?f=GPE%3D%281000%29%289.8%29%2830%29%3D2.94%5Ccdot%2010%5E5%20J)
The kinetic energy of the car is given by:
![KE=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where
m = 1000 kg
v = 50 m/s is the speed of the car
Substituting,
![KE=\frac{1}{2}(1000)(50)^2=1.25\cdot 10^6 J](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7D%281000%29%2850%29%5E2%3D1.25%5Ccdot%2010%5E6%20J)
Therefore, the mechanical energy is
![E=GPE+KE=2.94\cdot 10^5 + 1.25\cdot 10^6 = 1.54\cdot 10^6 J](https://tex.z-dn.net/?f=E%3DGPE%2BKE%3D2.94%5Ccdot%2010%5E5%20%2B%201.25%5Ccdot%2010%5E6%20%3D%201.54%5Ccdot%2010%5E6%20J)
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You are working for a manufacturing company, which is mathematically given as
<h3>What is the
value of m that will place the
movable bead in equilibrium at x-a a ....?</h3>
a)
Generally, the equation for the force of equilibrium is mathematically given as
F=2fcos\theta
Therefore
![K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3](https://tex.z-dn.net/?f=K%28npq%5E2%2Fa%5E2%29%3D2%5Cfrac%7Bkmpq%5E2%7D%7Ba%5Csqrt%7B2%7D%5E2%7D0.5%5C%5C%5C%5Cnp%3DmP%2F%20%5Csqrt%7B2%7D%5C%5C%5C%5Cwhere%20n%3D3)
![m=3\sqrt{2}](https://tex.z-dn.net/?f=m%3D3%5Csqrt%7B2%7D)
b)
By force equilibrium
![K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}](https://tex.z-dn.net/?f=K%28npq%5E2%2F%282a%5E2%29%29%3D2%2A%5Cfrac%7Bkmpq%5E2%7D%7Ba%5Csqrt%7B5a%7D%5E2%7D%2A%20%5Cfrac%7B2a%7D%7B%5Csart%7B5a%7D%7D)
Therefore
![n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\](https://tex.z-dn.net/?f=n%2F4%3D2%2F5%2Am%2A2%2F%5Csqrt%7B5%7D%5C%5C%5C%5Cm%3D%20%5Cfrac%7B5%5Csqrt%7B5%7D%7D%7B16%7D%5C%5C%5C%5C%5C%5C%5C%5C)
![m=\frac{15\sqrt{5}}{16}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B15%5Csqrt%7B5%7D%7D%7B16%7D)
c)
![K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}](https://tex.z-dn.net/?f=K%28npq%5E2%2Fx%5E2%29%3D2%5Cfrac%7Bkmpq%5E2%7D%7Ba%5Csqrt%7Bx%5E2%2Ba%5E2%7D%5E2%7D0.5%2Ax%2F%5Csqrt%7Bx%5E2%2Ba%5E2%7D)
x^2+a^2=(14/3)^{2/3}x^2
x=a/1.338
x=0.747a
d)
By force equilibrium
![K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}](https://tex.z-dn.net/?f=K%28npq%5E2%2Fx%5E2%29%3D2%5Cfrac%7Bkmpq%5E2n%7D%7B%5Csqrt%7Bx%5E2%2Ba%5E%7B3%2F2%7D%7D%5E2%7D%5C%5C%5C%5Cn%2Fx%5E2%3D%5Cfrac%7B2mx%7D%7B%28x%5E2%2Ba%5E2%29%5E%7B3%2F2%7D%7D)
![m/n=\frac{(x^2+a^2)3/2}{x^3}](https://tex.z-dn.net/?f=m%2Fn%3D%5Cfrac%7B%28x%5E2%2Ba%5E2%293%2F2%7D%7Bx%5E3%7D)
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The right answer is
2) NaCl
In fact, from the definition of acid, all acids contain at least one atom of hydrogen. In the options given by the problem, all compounds contain at least one atom of hydrogen H, except for the NaCl: therefore, NaCl is not an acid.
Answer:
![3.38\cdot 10^9 m](https://tex.z-dn.net/?f=3.38%5Ccdot%2010%5E9%20m)
Explanation:
The formula for the double-slit interference pattern is:
![y=\frac{m\lambda D}{d}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bm%5Clambda%20D%7D%7Bd%7D)
where:
m is the order of the maximum
is the wavelength
D is the distance between the slits and the screen
d is the distance between the two slits
If we assume for instance m=5, the formula gives us the distance between the fifth maximum and the first maximum. However, this will also gives us the distance between the fifth minimum and the first minimum, as the minima fall exactly between two maxima.
Therefore, if we use:
![m=5\\d=0.22 mm=2.2\cdot 10^{-4} m\\y=59 cm=0.59 m\\D=6.3 mm=6.3\cdot 10^{-3}m](https://tex.z-dn.net/?f=m%3D5%5C%5Cd%3D0.22%20mm%3D2.2%5Ccdot%2010%5E%7B-4%7D%20m%5C%5Cy%3D59%20cm%3D0.59%20m%5C%5CD%3D6.3%20mm%3D6.3%5Ccdot%2010%5E%7B-3%7Dm)
We can find the wavelength of the light:
![\lambda=\frac{yD}{md}=\frac{(0.59 m)(6.3\cdot 10^{-3} m)}{(5)(2.2\cdot 10^{-4} m)}=3.38 m=3.38\cdot 10^9 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7ByD%7D%7Bmd%7D%3D%5Cfrac%7B%280.59%20m%29%286.3%5Ccdot%2010%5E%7B-3%7D%20m%29%7D%7B%285%29%282.2%5Ccdot%2010%5E%7B-4%7D%20m%29%7D%3D3.38%20m%3D3.38%5Ccdot%2010%5E9%20m)