1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
7nadin3 [17]
3 years ago
15

Heat can be transferred through conduction, convection, and radiation. What is necessary in order for heat to be transferred by

conduction?
A.
Heat must be transferred through space.
B.
Heat must be transferred in the form of currents.
C.
The substances must be liquids or gases.
D.
The objects must be in direct contact with one another.
Physics
2 answers:
Pie3 years ago
7 0
Conduction is a mode of heat transfer where thermal energy flows between bodies that are in direct contact. Thus, the correct answer would be D.

A is incorrect here: heat flow via conduction involves high-energy particles (such as atoms or molecules) having their energy transferred to neighboring particles by contact or collision. If “space” refers to a region that contains only gas (the everyday understanding of “space”), then heat transfer by conduction would be extremely inefficient since gas molecules are so far apart. If “space” refers to a vacuum, then conduction cannot occur at all (radiation is the only mode of heat transfer through a vacuum). In either case, not only does conduction not necessitate space, but space impedes heat transfer by conduction.

B is incorrect because conduction operates via the movement of energy itself and not of matter (although, as we established above, the presence of matter is necessary for conduction). Conductive heat transfer can occur within a body itself (a metal bar that is heated on one end will eventually become hot on the other end). So, apart from being definitionally incorrect, conductive heat transfer doesn’t require currents at all. This mode of heat transfer occurs in convection.

C is manifestly incorrect: conductive heat transfer can occur between or within solids, so it can’t possibly be necessary that the “substances . . . Be liquids or gases.
jeka943 years ago
3 0

it would be..... C

sorry if I am wrong I tryed to think, At least I try!

You might be interested in
IF YOU ANSWER THESE 3 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!
gogolik [260]

Answer:

1. a

2. a [im iffy on this but 95% positive its this]

3. b [walking is a form of aerobics, so i would say b]

Explanation:

3 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
Which of the following changes would be a physical change to a substance?(1 point)
Naddika [18.5K]

Answer:

all of them I think

Explanation:

6 0
3 years ago
Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe
RideAnS [48]

Sarah's acceleration is -0.49 m/s^2

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

F_f = \mu mg

where

\mu is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

F=ma

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

F=-\mu mg = ma

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

a=-\mu g

And substituting the following values:

\mu = 0.05 (coefficient of friction)

g=9.81 m/s^2 (acceleration of gravity)

we find:

a=-(0.05)(9.81)=-0.49 m/s^2

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
Other questions:
  • Please help! Average speed. Show work!
    12·2 answers
  • An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 16.0 m if her initial speed is 3.60 m/
    13·1 answer
  • Why are bridges declared unsafe after a long use​
    14·1 answer
  • 30 points! Please answer at least one and how you found it! :)
    6·1 answer
  • Whats A Controlled Experiment? Whats A Controlled Experiment? Like what the definition is please!
    13·1 answer
  • Explain why a reference points that are stationary are usually chosen to determine whether an object is in motion.
    6·1 answer
  • At an equilibrium position of a pendulum, the <br> is at a maximum.
    15·1 answer
  • Which portion of the electromagnetic spectrum can the human eye detect? *
    5·1 answer
  • Pls someone help me
    7·1 answer
  • A turtle starts at a velocity of 2 m/s and accelerates at a rate of 3 m/s2. How long will it take the turtle to reach a final ve
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!