Use standard notation to represent the following elements. 1. potassium 2. copper 3. chlorine

an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe

yulyashka [42]

Answer:

The amount of electric charge transported = 0.192 C

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

Given: I = 285 mA, t = 674 milliseconds.

Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A

(ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s

Substituting these values into equation 1

Q = 0.285 × 0.674

Q = 0.192 C

Therefore the amount of electric charge transported = 0.192 C

5 0

2 years ago

An extended period when temperatures are well below average is known as a/an

Alecsey [184]

It would be a cold wave.
Ice ages occur when the temperatures are extremely cold.(long-term)

3 0

2 years ago

Read 2 more answers

A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene

Yuri [45]

Explanation:

The given data is as follows.

$U_{a} = 5.4 \times 10^{-8} J$

$W_{/text{a to b}} = -1.9 \times 10^{-8} J$

Electric potential energy ( $U_{b}$ ) = ?

Formula to calculate electric potential energy is as follows.

$U_{b} = U_{a} - W_{/text{a to b}}$

= $5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)$

= $7.3 \times 10^{-8} J$

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is $7.3 \times 10^{-8} J$ .

6 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

2 years ago

Use standard notation to represent the following elements. 1. potassium 2. copper 3. chlorine​

Use standard notation to represent the following elements. 1. potassium 2. copper 3. chlorine