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3 years ago

8

Get the lead out! That is really graphite in your pencil, not lead. Graphite is a form of carbon. Use the physical properties of

graphite to classify it using the choices below. A) metal B) nonmetal C) metalloid D) rare Earth element

2 answers:

Svetllana [295]3 years ago

5 0

NON-METAL!!!!!!!!!!!!!!!!!!!!!!!!!!!

Alja [10]3 years ago

4 0

The answer to this question is B. Nonmetal

You might be interested in

18. If the distance from a converging lens to the object is less than the focal length of the lens, the image will be _______ th

skad [1K]

Here's a quick way to find out. Pick up your glasses, bifocals work best, and find the focal length with a flashlight against a book. If I remember right, the object should be magnified and upside down. So, A.

5 0

3 years ago

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

kupik [55]

Answer:

a) $Q_{in} = 13.742\,kW$ , b) $\Delta S = 370.15\,\frac{kJ}{K}$

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

$Q_{in} +U_{sys,1} - U_{sys,2} = 0$

$Q_{in} = U_{sys,2} - U_{sys,1}$

$Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})$

$Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)$

$Q_{in} = 13.742\,kW$

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

$\Delta S = \frac{Q_{in}}{T_{in}}$

$\Delta S = \frac{13.742\,kJ}{370.15\,K}$

$\Delta S = 370.15\,\frac{kJ}{K}$

3 0

2 years ago

What do you want to learn about simple machines? (could someone come up with a question about simple machines for me)

bezimeni [28]

Answer:

what are simple machines lol

Explanation:

6 0

2 years ago

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel

insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t² (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t² (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t² (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

$p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125$

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0

2 years ago

Un prisma rectangular de cobre, de base igual a 36 centímetros cuadrados y una altura de 10 cm se sumerge hasta la mitad por med

butalik [34]

La respuesta correcta es 180 centímetros cúbicos o 180 $cm^{3}$

Explicación:

El primer paso para saber cuanto alcohol desaloja el prisma es calcular el volumen total del prisma. El volumen se puede encontrar usando la formula V (Volumen) = B (base) x h (altura). El proceso se muestra a continuación:

V = B x h

V = 36 $cm^{2}$ x 10 cm

V= 360 $cm^{3}$

Finalmente, el volumen total del prisma debe dividirse en 2 considerando que solo la mitad del prisma fue sumergida y esta mitad equivale al volumen del alcohol desplazado.

360 $cm^{3}$ ÷ 2 = 180 $cm^{3}$

5 0

2 years ago