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Artyom0805 [142]

3 years ago

11

An engineer is adding a heat sink to a motor to help absorb some of the heat produced by the motor. Which piece of metal would a

bsorb the MOST heat?

2 answers:

astraxan [27]3 years ago

6 0

<span>The metal that would absorb the MOST heat is Copper. </span>Copper is used in heat sink because it has many advantageous properties for thermally effective and long lasting heat converters..Copper is an excellent conductor of heat. It's high thermal conductivity allows heat to pass through it quickly. <span>Aluminium heat sinks are used as a cheaper, lightweight substitue to copper heat sinks, but they have a lower thermal conductivity than copper.</span>

soldi70 [24.7K]3 years ago

3 0

Answer:

2.0 kg of brass is the answer

You might be interested in

Please awnser and show the ways

topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0

3 years ago

A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch

ololo11 [35]

Answer:

$v_{1f}$ = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

p₀ = (m₁ + m₂) v₀

After the explosion

pf = m₁ $v_{1f}$ + m₂ $v_{2f}$

p₀ = [texpv_{f}[/tex]

(m₁ + m₂) v₀ = m₁ $v_{1f}$ + m₂ $v_{2f}$

Let's calculate the final speed (v1f) of the first stage

$v_{1f}$ = ((m₁ + m₂) v₀ - m₂ $v_{2f}$ ) / m₁

$v_{1f}$ = ((2100 +1160) 4300 - 1160 5940) / 2100

$v_{1f}$ = (14,018 10 6 - 6,890 106) / 2100

$v_{1f}$ = 7,128 106/2100

$v_{1f}$ = +3,394 103 m / s

come the same direction of the final stage, but more slowly

4 0

3 years ago

If the person drops box from 3.8 m how much energy is transferred from potential energy to kinetic energy

kotykmax [81]

Answer:

Kinetic energy

When work is done the energy is transferred from one type to another. This transferred energy may appear as kinetic energy.

For example, when you pedal your bicycle so that its speed increases, you are doing work to transfer chemical energy from your muscles to the kinetic energy of the bicycle.

Kinetic energy is the energy an object possesses by virtue of its movement. The amount of kinetic energy possessed by a moving object depends on the mass of the object and its speed. The greater the mass and the speed of the object the greater its kinetic energy.

The kinetic energy Ek of an object of mass m at a speed v is given by the relationship

{E_k} = \frac{1}{2}m{v^2}

m is the mass of the object in kilograms ( kg) and v is the speed of the object in metres per second ( m\,s^{-1}).

Explanation:

When work is done on an object it may also lead to energy being transferred to the object in the form of gravitational potential energy of the object.

Gravitational potential energy is the energy an object has by virtue of its position above the surface of the Earth. When an object is lifted, work is done. When work is done in raising the height of an object, energy is transferred as a gain in the gravitational potential energy of the object.

For example, suppose you lift a suitcase of mass m through a height h. The weight W of the suit case is a downward force of size mg. In lifting the suitcase, you would have to pull upwards on it with a force equal in size to its weight, mg.

Two suitcases. One has a green force arrow pointing up labelled F and a purple force arrow pointing down labelled 'Weight = mg'. The other case is raised by a height labelled h.

Suitcases with forces and height labelled

When this force (equal to the weight mg, but upwards) is applied to the suitcase over the distance h:

Work\,done=force\,\times\,distance\,upwards=mg\,\times\,h

This energy is transferred to potential energy when raising the object through a known height.

Energy = mass \times gravitational\,field\,strength \times height

E = m \times g \times h

This is the relationship used to calculate gravitational potential energy.

{E_p} = mgh

where m is the mass of the object in kilograms (kg), g is the gravitational field strength, (for positions near the surface of the Earth g = 9∙8 newtons per kilogram ( N kg ^{-1} and h is the height above the surface of the Earth in metres ( m).

8 0

3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

Diffusion occurs faster in gases than in liquids because _____.

jolli1 [7]

The correct answer to go in the blank would be A) The particles are moving faster.

5 0

3 years ago

Read 2 more answers