Answer:
I think its 9.0397 Ohms
Explanation:
take the reciprocal of all the resistances: 1/15, 1/65, 1/35
then add them: = 151/1365
then reciprocal the answer: =1365/151
And chuck it on a calculator: =9.04 Ohms
I think this is right but I'm not entirely sure. Tell me if I'm right by the way!
Answer:
W = 166422.729 N
Explanation:
given,
diameter of the balloon = 30 m
density of the air = 1.10 Kg/m³
weight of the balloon and cargo = ?
density of the surrounding air = 1.20 kg/m³
we know,
Density = mass/volume
m = density x volume
![m = \rho\times \dfrac{4}{3}\pi r^3](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%5Ctimes%20%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
![m =1.20 \times \dfrac{4}{3}\pi\times 15^3](https://tex.z-dn.net/?f=m%20%3D1.20%20%5Ctimes%20%5Cdfrac%7B4%7D%7B3%7D%5Cpi%5Ctimes%2015%5E3)
m = 16964.6 Kg
Weight of the balloon
W = m g
W = 16964.6 x 9.81
W = 166422.729 N
Weight of the balloon and the cargo is equal to W = 166422.729 N
B. they had different beliefs and backgrounds.
Because computers and technology develop and become faster things are going to become substantially easier.
Answer:
8.67×10⁻⁶ N/m²
Explanation:
p = Momentum of a photon
E = Energy of a photon
c = Speed of light
I = Intensity of light
Force = dp/dt
![p=\frac{E}{c}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7BE%7D%7Bc%7D)
![\\\Rightarrow F=\frac{\frac{dE}{dt}}{c}](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%20F%3D%5Cfrac%7B%5Cfrac%7BdE%7D%7Bdt%7D%7D%7Bc%7D)
As given in question
![F=2\frac{\frac{dE}{dt}}{c}](https://tex.z-dn.net/?f=F%3D2%5Cfrac%7B%5Cfrac%7BdE%7D%7Bdt%7D%7D%7Bc%7D)
Now F/A = Pressure
![P=\frac{2I}{c}\\\Rightarrow P=\frac{2\times 1300}{3\times 10^{8}}\\\Rightarrow P=8.67\times 10^{-6}\ N/m^2](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B2I%7D%7Bc%7D%5C%5C%5CRightarrow%20P%3D%5Cfrac%7B2%5Ctimes%201300%7D%7B3%5Ctimes%2010%5E%7B8%7D%7D%5C%5C%5CRightarrow%20P%3D8.67%5Ctimes%2010%5E%7B-6%7D%5C%20N%2Fm%5E2)
∴ Magnitude of the pressure on the sail is 8.67×10⁻⁶ N/m²