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3 years ago

Why are electrons, rather than protons, the principal charge carriers in metal wires?

1 answer:

6 0

<span>because in any atom the electrons are in the outer orbitals while protons are within the nucleus together with the neutrons. If energy is supplied electrons can jump to higher energy levels and leave the lower orbitals.
Especially in metals the conduction band is partially filled at room temperature with allows free flow of electrons throughout the metal thus they carry charge.
(it requires huge amounts of energy to remove a proton from the nucleus such things happen on the surface of sun).</span><span>
</span>

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A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

RSB [31]

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

3 0

3 years ago

Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-

torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0

3 years ago

If we increase the force applied to an object, and all other factors remain the same, the amount of work will

soldier1979 [14.2K]

The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.

-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.

-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.

-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.

3 0

3 years ago

Read 2 more answers

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

3 years ago

What effect does food colour have on the amount of food fish eat

Firlakuza [10]

Absolutely none. Fish take food into their mouths and taste it. If they like it, they swallow, if not they will spit it out. The color of the flakes has no bearing on what the fish takes in and tastes in my experience.
I hope this helps : )

8 0

3 years ago

Read 2 more answers