All categories

3 years ago

Which of the following best describes a plane?

Physics

1 answer:

ASHA 777 [7]3 years ago

7 0

From that list of choices, choice 'B' is the only example of a plane,
but it doesn't 'describe' it at all.

You might be interested in

A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p

agasfer [191]

Answer:

$\Delta P = 1961.4\,Pa$

Explanation:

The difference of pressure is given by gauge pressure:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)$

$\Delta P = 1961.4\,Pa$

8 0

3 years ago

Read 2 more answers

Part AIf the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negativ

Nana76 [90]

Answer:

Part a: The potential at point 3 and point 6 are V and 0 respectively.

Part b: The charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c: The net charge is 6CV.

Part d: The equivalent capacitance is 6C

Explanation:

As the question is not given here ,the complete question is found online and is attached herewith.

Part a:

V1 = V, V3 = V1 = V

and, V6 = V1-V = 0

The potential at point 3 and point 6 are V and 0 respectively

Part b

Q1 =CV = Q,

Q2 = 2C *V = 2Q

Q3 = 3 C*V = 3Q

So the charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c

Total charge of the system,

Q_net =Q1+Q2+Q3= (1+2+3) CV = 6 CV

So the net charge is 6CV.

Part d:

Equivalent Capacitance = Net charge / Voltage

Eq. C = 6CV/V = 6C

So the equivalent capacitance is 6C

6 0

3 years ago

I only need the answer to #21

Oksanka [162]

X-axis=time
y-axis=distance (position)

slope=y-axis / x-axis=distance / time=velocity.

answer: d) slope

6 0

3 years ago

A particle of mass m = 4.5 kg has velocity of v = 7 i m/s, when it is at the origin (0,0). Determine the z-component of the angu

Dima020 [189]

Answer:

Explanation:

Momentum P = Mass x Velocity

M = 4.5kg

V = 7m/s

4.5kg x 7m/s

= 31.5xkgm/s

L=IW( Angular momentum) at stationary origin (0,0)

I = 1/2 x Mr^2

L = 1/2 x 4.5x 31.5

L = 70.8kgm/s

At stationary point, (0,0) No coordinate exist

4 0

2 years ago

Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the

Leni [432]

Answer:

Part(a): The relative capacitance is $\dfrac{C_{A}}{C_{B}} = 0.33$

Part(b): The relative energy stored is $\dfrac{U_{A}}{U_{B}} = 0.33$

Part(c): The relative charge stored is $\dfrac{Q_{A}}{Q_{B}} = 0.33$

Explanation:

We know the capacitance ( $C$ ) of a capacitor having charge ( $Q$ ) and subjected to a potential difference of ( $V$ ) is given by

$C = \dfrac{Q}{V}$

Also, the energy ( $U$ ) stored by a capacitor can be written as

$U = \dfrac{1}{2}C~V^{2}$

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be $\textbf{r_{i}^{B}}$ $\bf{r_{i}^{B}}$ , the outer radius be $\bf{r_{o}^{B}}$ , the inner radius of Capacitor A be $\bf{r_{i}^{A}}$ and the outer radius be $\bf{r_{o}^{B}}$ .

Given in the problem,

$&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}$

Now, the capacitance ( $C$ ) of a cylindrical capacitor is given by,

$\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}$

where $\epsilon_{o}$ is the permittivity of the free space, $L$ is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

$C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}$

and the capacitance of capacitor B,

$C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}$