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2 years ago

What is a particulate ? Name a couple of examples.

1 answer:

3 0

Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles

Explanation:

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

2 years ago

What primary element of light helps the observer see three-dimensions?

elixir [45]

<span>Shading. When light hits an opaque surface some is absorbed, the rest is reflected, The reflected light is called shading. Reflection is not simple and varies with material. The surface’s structure defines the details of reflection. Variations produce anything from bright specular reflection</span>

7 0

3 years ago

Read 2 more answers

- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7

frez [133]

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b= -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

6 0

2 years ago

Why are eight electrons (four pairs) surrounding each non-hydrogen atom the optimal electronic arrangement for covalent molecule

rodikova [14]

Eight electrons surrounding each non-hydrogen atom is the optimal electronic arrangement for covalent molecules because it is needed to achieve an octet structure and is necessary to fill both the s and p subshells of electrons.

<h3>What is Covalent bonding?</h3>

This is the type of bonding which involves the sharing of electrons between atoms of an element.

This is done to achieve an octet configuration thereby making them stable and less reactive thereby making it the most appropriate choice.

Read more about Covalent bonding here brainly.com/question/3447218

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3 0

1 year ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$