Answer:
Magnitude of vector A = 0.904
Explanation:
Vector A , which is directed along an x axis, that is
![\vec{A}=x_A\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3Dx_A%5Chat%7Bi%7D)
Vector B , which has a magnitude of 5.5 m
![\vec{B}=x_B\hat{i}+y_B\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BB%7D%3Dx_B%5Chat%7Bi%7D%2By_B%5Chat%7Bj%7D)
![\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25](https://tex.z-dn.net/?f=%5Csqrt%7Bx_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%7D%3D5.5%5C%5C%5C%5Cx_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%3D30.25)
The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A ![\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%2B%5Cvec%7BB%7D%3D6x_A%5Chat%7Bj%7D%5C%5C%5C%5Cx_A%5Chat%7Bi%7D%2Bx_B%5Chat%7Bi%7D%2By_B%5Chat%7Bj%7D%3D6x_A%5Chat%7Bj%7D)
Comparing we will get
![x_A=-x_B\\\\y_B=6x_A](https://tex.z-dn.net/?f=x_A%3D-x_B%5C%5C%5C%5Cy_B%3D6x_A)
Substituting in ![x_{B}^{2}+y_{B}^{2}=30.25](https://tex.z-dn.net/?f=x_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%3D30.25)
![\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904](https://tex.z-dn.net/?f=%5Cleft%20%28-x_%7BA%7D%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%286x_%7BA%7D%20%5Cright%20%29%5E%7B2%7D%3D30.25%5C%5C%5C%5C37x_%7BA%7D%5E2%3D30.25%5C%5C%5C%5Cx_%7BA%7D%3D0.904)
So we have
![\vec{A}=0.904\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3D0.904%5Chat%7Bi%7D)
Magnitude of vector A = 0.904
Where are the images?!?!?
Answer:
a) ![k_{avg}=6.22\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D6.22%5Ctimes%2010%5E%7B-21%7D)
b) ![k_{avg}=8.61\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D8.61%5Ctimes%2010%5E%7B-21%7D)
c) ![k_{mol}=3.74\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D3.74%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
d) ![k_{mol}=5.1\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D5.1%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
Explanation:
Average translation kinetic energy (
) is given as
....................(1)
where,
k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K
T = Temperature in kelvin
a) at T = 27.8° C
or
T = 27.8 + 273 = 300.8 K
substituting the value of temperature in the equation (1)
we have
![k_{avg}=6.22\times 10^{-21}J](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D6.22%5Ctimes%2010%5E%7B-21%7DJ)
b) at T = 143° C
or
T = 143 + 273 = 416 K
substituting the value of temperature in the equation (1)
we have
![k_{avg}=8.61\times 10^{-21}J](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D8.61%5Ctimes%2010%5E%7B-21%7DJ)
c ) The translational kinetic energy per mole of an ideal gas is given as:
![k_{mol}=A_{v}\times k_{avg}](https://tex.z-dn.net/?f=k_%7Bmol%7D%3DA_%7Bv%7D%5Ctimes%20k_%7Bavg%7D)
here
= Avagadro's number; ( 6.02×10²³ )
now at T = 27.8° C
![k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D6.02%5Ctimes%2010%5E%7B23%7D%5Ctimes%206.22%5Ctimes%2010%5E%7B-21%7D)
![k_{mol}=3.74\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D3.74%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
d) now at T = 143° C
![k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D6.02%5Ctimes%2010%5E%7B23%7D%5Ctimes%208.61%5Ctimes%2010%5E%7B-21%7D)
![k_{mol}=5.1\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D5.1%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
Answer:
a) 39.6 m/s b) 4123 N
Explanation:
a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).
Fnet=ma
ma=m(v^2/R) (centripetal acceleration)
mg=m(v^2/R)
m cancels out (this is why pilot feels weightless) so,
g=(v^2/R)
9.8 m/s^2 = v^2/160 m
v^2=1568 m^2/s^2
v=39.6 m/s
b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.
Convert 300 km/hr to m/s
300 km/hr=83.3 m/s
Convert pilot's weight into mass:
760 N = 77.55 kg
Fnet=ma
n-mg=m(v^2/R)
n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)
n=3363.2 N+760 N=4123 N
In electricity, the most famous and basic equation is the Ohm's Law which relates the parameters voltage, current and resistance. One form of this law as written in equation is V = IR, where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. These parameters depends in the arrangements, whether it's series or parallel.
In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.