An image formed in a convex is smaller and the view is wider ture or false<br><br>

What are two ways in which ultrasound technology produces images

arsen [322]

Fetal and ophthalmic

6 0

3 years ago

Two large parallel metal plates are 1.2 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces

zheka24 [161]

Answer:

E=1100V/m

Explanation:

Given required <u>solution</u>

V=6.6v E=? V=Ed ; V is the potential difference between

d=D/2=1.2cm/2=0.6cm=0.006m the halfway

E is the electric field between the two

plates.

d is the distance between the halfway.

So we can use the above formula to calculate the electric field.

V=Ed from this E=V/d substitute the values from the given equation.

E=6.6v/0.006m

E= 1100 v/m

3 0

4 years ago

Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at

statuscvo [17]

$\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}$ $\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}$ $\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}$ $\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}$ $\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\ \end{gathered}$ $\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}$ Hence the correct answer is q= 1.73514 micro coulombs.