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2 years ago

Electrically charged particles are found primarily in

Physics

1 answer:

lesantik [10]2 years ago

8 0

Answer:

the ionosphere

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1. . The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?

natta225 [31]

Answer:In a DC circuit, the power consumed is simply the product of the DC voltage times the DC current, given in watts.for AC circuits with reactive components we have to calculate the consumed power differently.

a 1/4 watt resistor or a 20 watt amplifier.

3 0

3 years ago

There are many theories as to why ice ages occur. The currently accepted theory is that ice ages are caused by _____.

Harman [31]

Earlier in the text this question is provided for, you should be able to find the passage, "The currently accepted theory attributes ice ages to small changes in Earth's orbit, known as Milankovitch cycles. These cycles describe how Earth's orbit gradually changes shape from more circular to more elliptical and back again. This happens roughly every 100,000 years. The orbital changes take Earth nearer and further from the Sun, which would affect the amount of energy reaching Earth."

Answer: Small changes in Earth's orbit

7 0

3 years ago

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago

A 6.50-g sample of copper metal at 25.0°C is heated by the addition of 84.0 J of energy. The final temperature of the copper is

Mekhanik [1.2K]

Answer:

Final temperature of the copper is 59 degrees Celsius

Explanation:

It is given that,

Mass of the sample of copper metal, m = 6.5 g

Initial temperature of the metal, $T_i=25^{\circ}\ C=298\ K$

Heat generated, Q = 84 J

The specific heat capacity of liquid water is 0.38 J/g-K

Let $T_f$ is the final temperature of the copper. It can be calculated using the definition of specific heat of any substance. It is given by :

$Q=mc\Delta T$

$Q=mc(T_f-T_i)$

$T_f=\dfrac{Q}{mc}+T_i$

$T_f=\dfrac{84}{6.5\times 0.38}+298$

$T_f=332\ K$

$T_f=59^{\circ}C$

So, the final temperature of the copper is 59 degrees Celsius. Hence, this is the required solution.

4 0

3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$