Anything that has volume or mass

Which of the following is false regarding the structure of atoms?

Ad libitum [116K]

Hmmm. I think the answer is A. If I remember correctly. I had this question before. Sorry if I'm wrong.

6 0

4 years ago

100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.

FinnZ [79.3K]

A) The power delivered to the lines is
$P_{in}= 100 kW=1 \cdot 10^5 W$
And the voltage at which the lines work is
$V=13014.1 V$
Since the power delivered is the product between the voltage and the current:
$P=VI$
We can find the current flowing in the lines:
$I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A$

b) The voltage change along each line can be found by using Ohm's law:
$\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V$

c) The power wasted as heat along each line is given by:
$P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W$
And since we have 2 lines, the total power wasted as heat in both lines is
$P_d = 2 \cdot 590 W=1180 W$

6 0

3 years ago

What is rickrolling<br><br> ...

Andreyy89

Answer:

"a prank and an Internet meme involving an unexpected appearance of the music video for the 1987 Rick Astley song "Never Gonna Give You Up". The meme is a type of bait and switch using a disguised hyperlink that leads to the music video."

Explanation: G O O G L E

8 0

3 years ago

Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, t

Reil [10]

Answer:

a = 16.5 x 10⁻⁶ m = 16.5 μm

Explanation:

Here we will use the diffraction equation:

y = mλL/a

where,

y = distance between two consecutive dark fringes = 5.34 cm = 0.0534 m

m = order of diffraction = 1

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

L = Distance between hair and screen = 1.4 m

a = thickness of hair = ?

0.0534 m = (1)(6.33 x 10⁻⁷ m)(1.4 m)/(a)

a = (6.33 x 10⁻⁷ m)(1.4 m)/(0.0534 m)

5 0

3 years ago

When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 79 % of the dielectr

Oksanka [162]

Complete question:

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 × 10⁷ V/m . A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates.

a- When the electric field between the plates is 79 % of the dielectric strength, what is the energy density of the stored energy?

b- When the capacitor is connected to a battery with voltage 500.0 V , the electric field between the plates is 79 % of the dielectric strength. What is the area of each plate if the capacitor stores 0.25 mJ of energy under these conditions?

Answer:

(a) The energy density of the stored energy is 2872.1 J/m³

(b) Area of the plate is 27.51 cm²

Explanation:

Given;

dielectric constant, K = 2.6

dielectric strength = 2.0 × 10⁷ V/m

potential difference of the battery, V = 500 V

Part (a)

Electric field strength, E = 79 % of dielectric strength

E = 0.79 x 2.0 × 10⁷ V/m

E = 1.58 × 10⁷ V/m

The energy density of the stored energy

u = ¹/₂Kε₀E²

where;

u is the energy density

ε₀ is permittivity of free space = 8.85 x 10⁻¹² C²/N.m²

u = ¹/₂Kε₀E² = ¹/₂ x 2.6 x 8.85 x 10⁻¹² x (1.58 × 10⁷)²

u = 28.721 x 10² J/m³

u = 2872.1 J/m³

Part (b) Area of the plate if the capacitor stores 0.250 mJ of energy

$A = \frac{C_oV}{\epsilon_o E}$

The energy density of stored energy is also given as;

U = ¹/₂KC₀V², U = 0.250 mJ

The charge stored in the capacitor;

$C_o =\frac{2U}{KV^2} \\\\C_o =\frac{2*0.25*10^{-3}}{2.6*(500)^2} = 7.692 *10^{-10} = 0.7692 \ nF$

Substitute these values and solve for Area

$A = \frac{C_oV}{\epsilon_o E} = \frac{7.692*10^{-10} *500}{8.85*10^{-12}*1.58*10^7} = 2.751*10^{-3} \ m^2$

A = 27.51 cm²

3 0

4 years ago