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4 years ago

I need to present a proposal for a device that minimizes heat transfer. I need help PLEASE

1 answer:

3 0

Engineers must understand the thermal properties of materials to be able to predict the performance of any given material over its lifetime in a specific application. Engineers apply their understanding of the thermal properties of materials to the design of efficient heat transfer materials for better engines, spacecraft and electronic devices. They also examine the thermal properties of insulation to design more efficient buildings and homes. Engineers develop ways to minimize heat transfer from a motor to the surrounding environment. Often they find ways to insulate the motor to decrease the convective heat transfer from the motor. They design a refrigerator to keep heat out of the inside, as well as keep the refrigerator contents cool. hope this helped

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AYUDA!!!!!

ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 $\frac{g}{cm^{3} }$ o 790 $\frac{g}{cm^{3} }$

b) El peso especifico es 7749.9 $\frac{N}{m^{3} }$

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

$d=\frac{m}{V}$

En este caso:

masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

$d=\frac{237 g}{300cm^{3} }$ → d=0.79 $\frac{g}{cm^{3} }$

$d=\frac{0,237 kg}{0,0003m^{3} }$ → d=790 $\frac{g}{cm^{3} }$

El valor de la densidad es 0.79 $\frac{g}{cm^{3} }$ o 790 $\frac{g}{cm^{3} }$

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 $\frac{m}{s^{2} }$ = 2,32497 N

el peso especifico es calculado como:

$Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }$

Pe= 7749.9 $\frac{N}{m^{3} }$

El peso especifico es 7749.9 $\frac{N}{m^{3} }$

8 0

3 years ago

Conduction is heat in solids and liquids by _______

elena55 [62]

The answer is gases.

hope this help

4 0

3 years ago

A small block slides down an incline with a constant acceleration. The block is released from rest at the top of the incline. Af

zloy xaker [14]

Answer:

v = 3.24 m/s

Explanation:

Since we don't have time, we can use the formula;

(Final distance - initial distance)/time = (initial velocity + final velocity)/2

Thus;

(x_f - x_i)/t = ½(v_xi + v_xf)

We are given;

x_i = 0 m

x_f = 5 m

v_xi = 0 m/s

v_xf = 5 m/s

Thus, plugging in the relevant values;

(5 - 0)/t = (0 + 5)/2

5/t = 5/2

t = 2 s

Using Newton's first law of motion, we can find the acceleration.

v = u + at

Applying to this question;

5 = 0 + a(2)

5 = 2a

a = 5/2

a = 2.5 m/s²

To get the speed, in m/s, of the block when it had traveled only 2.1 m from the top, we will use the formula;

v² = u² + 2as

v² = 0² + 2(2.5 × 2.1)

v² = 10.5

v = √10.5

v = 3.24 m/s

3 0

4 years ago

A sanding disk with rotational inertia 3.8 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnit

Elis [28]

Answer:

(a) Angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b) Angular velocity of the disk is $353\frac{rad}{s}$

Explanation:

Given

Rotational inertia of the disk , $I=3.8\times 10^{-3}kg.m^{2}$

Torque delivered by the motor , $\tau =17N.m$

Torque is applied for duration , $\Delta t=79ms=0.079s$

(a)

Magnitude of angular momentum of the disk = Angular impulse produced by the torque

$\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}$

=> $L=1.343\, \frac{kg.m^{2}}{s}$

Thus angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b)

Since Angular momentum , $L=I\omega$

where $\omega$ = Angular velocity of the disk

$=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}$

$\therefore \omega =353\frac{rad}{s}$

Thus angular velocity of the disk is $353\frac{rad}{s}$

5 0

4 years ago

What is the change in velocity of a 15 kg object that experiences a 5 N force for 0.3 seconds?

alukav5142 [94]

Hello!

Δv = 0.1 m/s

Use the equation F = m · a to solve for the acceleration:

5 = 15 · a

a = 1/3

The equation to solve for acceleration can be rewritten to solve for the change in velocity:

Δv / t = a

Δv = at

We are given the acceleration and time, therefore:

Δv = (1/3)(0.3)

Δv = 0.1 m/s

3 0

3 years ago