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2 years ago

11

Which statement correctly describes why a compound is a pure substance?

1 answer:

ad-work [718]2 years ago

6 0

Explanation:

A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.

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A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

3 years ago

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

A

$x = 0.198456 \ m$

B

$h = 1.3061 \ m$

C

$v = 5.06 \ m/s$

D

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

5 0

2 years ago

A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 3.84 m/s

JulijaS [17]

Answer:

the train is moving at the speed of v = 1.79 m/s

Explanation:

given,

rain drop is falling vertically down with the speed of = 3.84 m/s

angle of the rain drop = 25°

tan θ = $\dfrac{v}{u}$

tan 25° = $\dfrac{v}{3.84}$

v =3.84 × tan 25°

v = 1.79 m/s

hence, the train is moving at the speed of v = 1.79 m/s

3 0

2 years ago

Ways that charges can be transferred

Rama09 [41]

Can I have more Details?

8 0

2 years ago

Read 2 more answers

Given two vectors A=4i^+3j^ and vector<br> B=5i^-2j^.find the magnitude of each vector

Flura [38]

Answer:

<em>Magnitude of A=5</em>

<em>Magnitude of B=5.39</em>

Explanation:

<u>The magnitude of Vectors in Rectangular Form</u>

Given a vector v in its rectangular form:

$\mathbf{v}=x\hat i+y\hat j$

The magnitude of v is:

$\mid\mid\mathbf{v}\mid \mid=\sqrt{x^2+y^2}$

We are given the vectors

$\mathbf{A}=4\hat i+3\hat j$

$\mathbf{B}=5\hat i-2\hat j$

Their magnitudes are:

$\mid\mid\mathbf{A}\mid \mid=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}$

$\mid\mid\mathbf{A}\mid \mid=5$

$\mid\mid\mathbf{B}\mid \mid=\sqrt{5^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}$

$\mid\mid\mathbf{B}\mid \mid=\sqrt{29}=5.39$

4 0

2 years ago