Ask question

Ask question

All categories

4 years ago

7

3. Describe the process of heat transfer by conduction. What happens to the molecules that cause the thermal energy to be transf

erred?

1 answer:

nasty-shy [4]4 years ago

6 0

What are the answer choices?

You might be interested in

Meteoric material dates the formation of the solar system at about_____ billion years.

swat32

Meteoric material dates the formation of the solar system at about <u>4.6 billion</u> years ago

The solar system was formed from a dense cloud of interstellar gas and dust called the solar nebula about 4.53 and 4.58 billion years ago. It could collapse, possibly due to the shockwave of a nearby exploding star called supernova.

<h3>What is the solar system?</h3>

The solar system is the gravitationally bound system of the sun, the objects that orbit it or travel around the sun. It consists of an average star (sun), the planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto) and their moons, dwarf planets and countless asteroids, comets, and other small icy objects.

Learn more about solar system at brainly.com/question/14866884

#SPJ4

4 0

2 years ago

Mark and Paul are in a race. Mark is 20 meters from the finish line and running at a constant 3.5 m/s. Paul is 5 meters behind h

USPshnik [31]

Answer:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

Explanation:

For this case we have an illustration for the problem on the figure attached.

And we can solve this problem analyzing each one of the runner. Let's begin with Mark

Mark

For this case we know that $V_M = 3.5 m/s$ and th velocity is constant. The distance from Mark and the finish line is $D_M = 20 m$

Since the velocity is constant we can create the following relation:

$D_M = V_M t_M$

And solving for $t_M$ we got:

$t_m = \frac{D_M}{V_M}= \frac{20m}{3.5 m/s}= 5.71 s$

So then Mark will nd the race after 5.71 seconds

Paul

We know that the initial velocity for Paul is given $V_{iP}= 2.7 m/s$ we also know that the total distance between Paul and the finish line is 25 m and we want to find the acceleration that Paul needs to apply in order to tie the race, and Paul have 5.71 sconds in order to reach the finish line.

We can use this formula in order to find the acceleration (because we assume that the acceleration is constant) that he needs to apply:

$x_f = x_i + v_i t + \frac{1}{2} a t^2$

And since $\Delta x = x_f - x_i$ we have this:

$\Delta x= v_i t + \frac{1}{2} a t^2$

And if we replace we have this:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

And the final velocity for Paul using this acceleration would be:

$V_{fP}= V_{iP}+ a_P t = 2.7m/s + 0.588 m/s^2 (5.71s)= 6.057 m/s$

3 0

4 years ago

What mass of steam at 100°C must be mixed with 490 g of ice at its melting point, in a thermally insulated container, to produce

sukhopar [10]

Answer:

the mass of steam at 100°C must be mixed is 150 g

Explanation:

given information:

ice's mass, $m_{i}$ = 490 g = 0.49 kg

steam temperature, T = 100°C

liquid water temperature, T = 89.0°C

specific heat of water, $c_{w}$ = 4186 J/kg.K = 4.186 kJ/kg.K

latent heat of fusion, $L_{f}$ = 333 kJ/kg

latent heat of vaporization, $L_{v}$ = 2256 kJ/kg

first, we calculate the heat of melted ice to water

Q₁ = $m_{i} L_{f}$

where

Q = heat

$m_{i}$ = mass of the ice

$L_{f}$ = latent heat of fusion

thus,

Q₁ = $m_{i} L_{f}$

= 0.49 x 333

= 163.17 kJ

then, the heat needed to increase the temperature of water to 89.0°C

Q₂ = $m_{i}$ $c_{w}$ (89 - 0), the temperature of ice is 0°C

$c_{w}$ = specific heat of water

so,

Q₂ = $m_{i}$ $c_{w}$ (89 - 0)

= 0.49 x 4.186 x 89

= 182.55 kJ

so, the heat absorbed by the ice is

Q = Q₁ + Q₂

= 163.17 + 182.55

= 345.72 kJ

the temperature of the steam is 100°C, so the mass of the steam is

Q = $m_{s}$ $L_{v}$ + $m_{s}$ $c_{w}$ (100 - 89)

Q = $m_{s}$ ( $L_{v}$ + $c_{w}$ (11))

$m_{s}$ = Q/ [ $L_{v}$ + $c_{w}$ (11)]

= 345.72/ [2256 + (4.186 x 11)]

= 0.15 kg

= 150 g

5 0

3 years ago

Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

3 0

3 years ago

Choose the correct statement below that accurately describes the shear and normal stresses in a beam. A. Shear stresses are maxi

Scorpion4ik [409]

Answer:

A. Shear stresses are maximum at the neutral axis and normal stresses are maximum furthest from the neutral axis.

Explanation:

Normal stress :

Normal stress is defined as the stress or the restoring force that occurs on the plane when an external axial load is applied on it. For a beam the normal stress is maximum at the point furthest from the neutral axis and is zero at the neutral axis of the beam.

Shear stress :

Shear stress is a stress which occurs when the force acts on the surface of the member in a parallel direction. It changes the shape of the member. For a beam, the shear stress is maximum at the neutral axis.

6 0

3 years ago