Answer:
Maximum speed of the car is 17.37 m/s.
Explanation:
Given that,
Radius of the circular track, r = 79 m
The coefficient of friction, ![\mu=0.39](https://tex.z-dn.net/?f=%5Cmu%3D0.39)
To find,
The maximum speed of car.
Solution,
Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :
![v=\sqrt{\mu rg}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cmu%20rg%7D)
![v=\sqrt{0.39\times 79\times 9.8}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B0.39%5Ctimes%2079%5Ctimes%209.8%7D)
v = 17.37 m/s
So, the maximum speed of the car is 17.37 m/s.
Answer:
The relationship is only between the coefficients A, E and J which is:
. The remaining coefficients can be anything without any constraints.
Explanation:
Given:
The three components of velocity is a velocity field are given as:
![u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz](https://tex.z-dn.net/?f=u%20%3D%20Ax%20%2B%20By%20%2B%20Cz%5C%5C%5C%5Cv%20%3D%20Dx%20%2B%20Ey%20%2B%20Fz%5C%5C%5C%5Cw%20%3D%20Gx%20%2B%20Hy%20%2B%20Jz)
The fluid is incompressible.
We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,
![\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%2B%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%2B%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D0)
Now, let us find the partial derivative of each component.
![\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%28Ax%2BBy%2BCz%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3DA%2B0%2B0%3DA%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D%28Dx%2BEy%2BFz%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%3D0%2BE%2B0%3DE%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20z%7D%28Gx%2BHy%2BJz%29%5C%5C%5C%5C%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D0%2B0%2BJ%3DJ)
Hence, the relationship between the coefficients is:
![A+E+J=0](https://tex.z-dn.net/?f=A%2BE%2BJ%3D0)
There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.
Answer:
The focal length of the mirror is 52.5 cm.
Explanation:
Given that,
Object to Image distance d = 140 cm
Image distance v= 35 cm
We need to calculate the object distance
![u = d-v](https://tex.z-dn.net/?f=u%20%3D%20d-v)
![u = 140-35=105\ cm](https://tex.z-dn.net/?f=u%20%3D%20140-35%3D105%5C%20cm)
We need to calculate the focal length
Using formula of mirror
![\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bu%7D%2B%5Cdfrac%7B1%7D%7Bv%7D)
Put the value into the formula
![\dfrac{1}{f}=\dfrac{1}{-105}+\dfrac{1}{35}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7B-105%7D%2B%5Cdfrac%7B1%7D%7B35%7D)
![\dfrac{1}{f}=\dfrac{2}{105}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B2%7D%7B105%7D)
![f=52.5\ cm](https://tex.z-dn.net/?f=f%3D52.5%5C%20cm)
Hence, The focal length of the mirror is 52.5 cm.
Answer:
7) λ = 0.5 m, 8) f = 4.8 10¹⁴ Hz
Explanation:
The speed of an electromagnetic wave is
c = λ f
where c is the speed of light in vacuum c = 3 10⁸ m / s
7) indicate the frequency f = 6.0 10⁸ Hz
we do not know the wavelength
λ = c / f
we calculate
λ = 3 10⁸ / 6.0 10⁸
λ = 0.5 m
8) indicate the wavelength λ = 6.25 10-7 m
we do not know the frequency
f = c / λ
we calculate
f = 3 10⁸ / 6.25 10⁻⁷
f = 0.48 10¹⁵ Hz
f = 4.8 10¹⁴ Hz
Answer:
Transfer of charge by touching is called conduction.