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Marysya12 [62]
4 years ago
5

Heat transfer between two substances is affected by specific heat and the

Physics
1 answer:
IgorLugansk [536]4 years ago
6 0

The correct answer is D. Amount of time and area of physical contact between the substances.

Explanation:

Heat transfer refers to the flow of thermal energy or heat between two or more objects. This process involves multiple factors and implies heat from the hottest object goes to the coldest one until there is an equilibrium. To begin, heat transfer depends on the amount of thermal energy in the objects because objects must have a different amount of thermal energy for heat to flow.

Besides this, the amount of energy that flows depends on the time and the contact between the substances of objects. Indeed, objects need to be in contact or close to each other for heat to transfer, and the time needs to be enough for the process to occur. For example, if you place a pot over the fire just for a few seconds it is likely the heat transferred is minimal, which does not occur if you leave the pot more time. At the same time if the pot is in close contact with fire more heat will be transferred.-

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How to represent milligram in kilogram by standard formula?
Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take   \frac{1}{1,000,000} which equals 0.000001 kg

4 0
3 years ago
What is the test for whether or not a moving object is in equilibrium?
Sphinxa [80]

It is in equilibrium if its velocity is not changing.

3 0
3 years ago
Read 2 more answers
The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter yo
Alexus [3.1K]

Answer:

k_2 = 7.815 * 10^-3 s^-1

Explanation:

Given:

- rate constant of reaction k_1 = 7.8 * 10^-3 s^-1    @  T_1 = 25 C

- rate constant of reaction k_2 = ?    @  T_2 = 75 C

- The activation energy E_a = 33.6 KJ/mol

- Gas constant R = 8.314472 KJ / mol . K

Find:

- rate of reaction k_2 @ T_2 = 75 C

Solution:

- we will use a combined form of Arrhenius equations that relates rate constants k as function of E_a and temperatures as follows:

                              k_2 = k_1 * e ^ [(E_a / R) * ( 1 / T_1 - 1 / T_2 )

- Evaluate             k_2 = 7.8 * 10^-3* e^[(33.6 / 8.314472)*(1/298 -1/348)

- Hence,                k_2 = 7.815 * 10^-3 s^-1

4 0
4 years ago
Which type of electronic community allows real-time discussion among members?
Brilliant_brown [7]

Instant messaging is the type of electronic community that allows real-time discussion among members.

 

Instant messaging<span> <span>(IM) is a type of online chat which offers real-time text transmission over the Internet. A LAN </span></span>messenger<span> <span>operates in a similar way over a local area network. </span></span>

 

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

4 0
4 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
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