Answer:
P = 1471500 [Pa]
Explanation:
We must remember that pressure is defined as the relationship between Force over the area.
![P=F/A](https://tex.z-dn.net/?f=P%3DF%2FA)
where:
P = pressure [Pa] (units of pascals)
F = force [N] (units of Newtons)
A = area of contact = 4 [cm²]
But first we must convert from cm² to m²
![A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }](https://tex.z-dn.net/?f=A%20%3D%204%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B100%5E%7B2%7D%20cm%5E%7B2%7D%20%7D)
A = 0.0004 [m²]
Also, the weight should be calculated as follows:
![w = m*g](https://tex.z-dn.net/?f=w%20%3D%20m%2Ag)
where:
m = mass = 60 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
![w = 60*9.81\\w = 588.6[N]](https://tex.z-dn.net/?f=w%20%3D%2060%2A9.81%5C%5Cw%20%3D%20588.6%5BN%5D)
And the pressure:
![P=588.6/0.0004\\P=1471500 [Pa]](https://tex.z-dn.net/?f=P%3D588.6%2F0.0004%5C%5CP%3D1471500%20%5BPa%5D)
Because 1 [Pa] = 1 [N/m²]
Answer:
With changing speed and/or in a circle
Answer:
Current (I) = 3 x 10^-2 A
Explanation:
As we know, ![B = 4\pi 10^-7 *l/ 2\pi r](https://tex.z-dn.net/?f=B%20%3D%204%5Cpi%2010%5E-7%20%2Al%2F%202%5Cpi%20r)
By putting up the values needed from the data...
Current (I) = 2 x 3.14 x (3.0 x 10^-6) (2.0 x 10^-3) / 4 x 3.14 x 10^-7 = 3 x 10^-2 A
The question is incomplete. The complete question is :
A plate of uniform areal density
is bounded by the four curves:
![$y = -x^2+4x-5m$](https://tex.z-dn.net/?f=%24y%20%3D%20-x%5E2%2B4x-5m%24)
![$y = x^2+4x+6m$](https://tex.z-dn.net/?f=%24y%20%3D%20x%5E2%2B4x%2B6m%24)
![$x=1 \ m$](https://tex.z-dn.net/?f=%24x%3D1%20%5C%20m%24)
![$x=2 \ m$](https://tex.z-dn.net/?f=%24x%3D2%20%5C%20m%24)
where x and y are in meters. Point
has coordinates
and
. What is the moment of inertia
of the plate about the point
?
Solution :
Given :
![$y = -x^2+4x-5$](https://tex.z-dn.net/?f=%24y%20%3D%20-x%5E2%2B4x-5%24)
![$y = x^2+4x+6$](https://tex.z-dn.net/?f=%24y%20%3D%20x%5E2%2B4x%2B6%24)
![$x=1 $](https://tex.z-dn.net/?f=%24x%3D1%20%24)
![$x=2 $](https://tex.z-dn.net/?f=%24x%3D2%20%24)
and
,
,
.
So,
![$dI = dmr^2$](https://tex.z-dn.net/?f=%24dI%20%3D%20dmr%5E2%24)
, ![$r=\sqrt{(x-1)^2+(y+2)^2}$](https://tex.z-dn.net/?f=%24r%3D%5Csqrt%7B%28x-1%29%5E2%2B%28y%2B2%29%5E2%7D%24)
![$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$](https://tex.z-dn.net/?f=%24dI%20%3D%20%28%5Crho%29%28%28x-1%29%5E2%2B%28y%2B2%29%5E2%29dx%20%5C%20dy%24)
![$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$](https://tex.z-dn.net/?f=%24I%3D%202%20%5Cint_1%5E2%20%5Cint_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%28%28x-1%29%5E2%2B%28y%2B2%29%5E2%29%20dy%20%5C%20dx%24)
![$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$](https://tex.z-dn.net/?f=%24I%3D%202%20%5Cint_1%5E2%20%5Cint_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%28x-1%29%5E2%2B%28y%2B2%29%5E2%20%5C%20%20dy%20%5C%20dx%24)
![$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$](https://tex.z-dn.net/?f=%24I%3D2%20%5Cint_1%5E2%20%5Cleft%28%20%5Cleft%5B%20%28x-1%29%5E2y%2B%5Cfrac%7B%28y%2B2%29%5E3%7D%7B3%7D%5Cright%5D_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%5Cright%29%20%5C%20dx%24)
![$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$](https://tex.z-dn.net/?f=%24I%3D2%20%5Cint_1%5E2%20%28x-1%29%5E2%20%282x%5E2%2B11%29%2B%5Cfrac%7B1%7D%7B3%7D%5Cleft%28%28x%5E2%2B4x%2B6%2B2%29%5E3-%28-x%5E2%2B4x-5%2B2%29%5E3%20%5C%20dx%24)
![$I=\frac{32027}{21} \times 2$](https://tex.z-dn.net/?f=%24I%3D%5Cfrac%7B32027%7D%7B21%7D%20%5Ctimes%202%24)
![$= 3050.19 \ kg \ m^2$](https://tex.z-dn.net/?f=%24%3D%203050.19%20%5C%20kg%20%5C%20m%5E2%24)
So the moment of inertia is
.
Answer:
The potential difference between the plates is 596.2 volts.
Explanation:
Given that,
Capacitance ![C=260\ pF](https://tex.z-dn.net/?f=C%3D260%5C%20pF)
Charge ![q=0.155\ \mu\ C](https://tex.z-dn.net/?f=q%3D0.155%5C%20%5Cmu%5C%20C)
Separation of plates = 0.313 mm
We need to calculate the potential difference between the plates
Using formula of potential difference
![V= \dfrac{Q}{C}](https://tex.z-dn.net/?f=V%3D%20%5Cdfrac%7BQ%7D%7BC%7D)
Where, Q = charge
C = capacitance
Put the value into the formula
![V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B0.155%5Ctimes10%5E%7B-6%7D%7D%7B260%5Ctimes10%5E%7B-12%7D%7D)
![V=596.2\ volts](https://tex.z-dn.net/?f=V%3D596.2%5C%20volts)
Hence,The potential difference between the plates is 596.2 volts.