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ra1l [238]
3 years ago
12

The difference between the speed of sound n air at 0°C and the speed of sound in air at 20°c is that... A. Cooler air molecules

move more slowly and respond less readily to the energy of a sound wave. B. Cooler air molecules move more quickly and respond less readily to the energy of a sound wave. C. Cooler air molecules do not move and respond less readily to the energy of a sound wave. D. Cooler air molecules move more randomly and respond less readily to the energy of a sound wave.
Physics
2 answers:
sergejj [24]3 years ago
7 0
Heat, like sound, is kinetic energy. Molecules at higher temperatures heave more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly. 

So the answer is A.
KiRa [710]3 years ago
3 0

Answer:

on usatest prep its b

Explanation:

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Define project torque
pogonyaev

Answer:

ExplanProject Torque was the free North American version of Level-R. It is a multiplayer online racing game (MMORG) with partially chargeable content, or micro-transactions. It features gameplay elements such as tuning and customization. The game is titled as Level R in Europe with a slightly different game interface and menus. Europe, US, Russia, Indonesia, Thailand, China, and Japan versions, are developed by Invictus Games and are subject to Invictus copyrightation:

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3 years ago
Hee Sun drew an electron dot diagram of a silicon atom as shown. In addition to changing the symbol to C, how would this diagram
dolphi86 [110]
Other than for the chemical symbol, the electron dot diagram for silicon would be the same as it was for carbon.

The reason for this is because electron dot diagrams are used to represent the electrons in the outermost, or valence, shell of an atom. In a group of the periodic table, all of the elements have the same number of valence shell electrons. This means that all elements belonging to the same group have the same electron dot diagram, except for the symbol of the element that is within the diagram.
8 0
3 years ago
As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile
Ber [7]

Answer:

a)\Delta T= 100^{\circ}C

b)\bigtriangledown T=1^{\circ}C.mile^{-1}

c)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

d)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

T=(d+30) ^{\circ}C...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

\Delta T= T_f-T_i..............................(2)

where:

T_f= final temperature

T_i= initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

\Delta T= (100+30)-(0+30)

\Delta T= 100^{\circ}C

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

\bigtriangledown T=\frac{\Delta T}{\Delta d}......................(3)

where:

\Delta d= change in distance

\bigtriangledown T=change in temperature with respect to distance

putting the respective values in eq. (3)

\bigtriangledown T=\frac{100}{100}

\bigtriangledown T=1^{\circ}C.mile^{-1}

(c)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

(d)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

4 0
3 years ago
How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m
Mars2501 [29]

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

Initial speed, u = 0

Final speed, v = 25 m/s

Distance, d = 200 m

We need to find the net force. Firstly, let's find acceleration using equation of motion.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2

Net force, F = ma

F=1600\times 1.56\\\\F=2496\ N

So, the net force is 2496 N.

6 0
3 years ago
Convert the following.
Alexandra [31]

Answer:

a-

50°C whn converted into K - 323 K

4 0
3 years ago
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