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
NARA [144]
2 years ago
7

Temperature refers to the _____. A. amount of heat energy in an object B. arrangement of the atoms in matter C. average kinetic

energy of the particles in a substance D. pressure exerted by atoms on anything that touches them
Physics
1 answer:
lawyer [7]2 years ago
8 0
I believe A. is the answer. Temperature is the measure of heat, heat is considered to be the amount of vibration within an atom. Hot atoms will vibrate a lot, cooler atoms will no vibrate much. You can heat atoms up and cause them to break their bonds with other atoms and create liquids or gases. Hence, you can cool atoms and cause them to revert into a liquid from a gas, or solid from a liquid. 
You might be interested in
A brick sliding across a smooth floor has a coefficient of static friction of s=0.23 and a coefficient of kinetic friction of k=
pantera1 [17]

Answer:

0.029325

Explanation:

0.23x0.15=0.0345

0.0345x0.85=0.029325

4 0
1 year ago
Read 2 more answers
What is the relationship between frequency ad wavelenght
Law Incorporation [45]

Answer:

Frequency and wavelength are inversely proportional to each other. The wave with the greatest frequency has the shortest wavelength. Twice the frequency means one-half the wavelength. For this reason, the wavelength ratio is the inverse of the frequency ratio.

7 0
3 years ago
Read 2 more answers
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
How are pupae, larvae, and nymphs similar? *Answer Fast!*
Goshia [24]

Answer:

They are all a cycle!

Explanation:

They just are all cycles.

5 0
3 years ago
When using a spring scale, the measurements you obtain will be in ____.
SpyIntel [72]
The answer is letter c

6 0
3 years ago
Read 2 more answers
Other questions:
  • Nami conducts an investigation on plants. She places a grow light on a timer to give the plants different amounts of light to se
    9·2 answers
  • What is the distance between two points (3,4,-5) and (2,1,0,)
    14·1 answer
  • Two violinists are playing their "A" strings. Each is perfectly tuned at 440 Hz and under 245 N of tension. If one violinist tur
    14·1 answer
  • In an extreme marathon, participants run a total of 100km; world-class athletes maintain a pace of 15 km/h. how many 230 Calorie
    12·2 answers
  • An object has a relativistic momentum that is 8.30 times greater than its classical momentum. What is its speed?
    7·2 answers
  • The amount of force needed to keep a 0.5 kg football moving at a constant speed of 8 m/s
    11·1 answer
  • A marble statute is exposed to the weather, within a few years
    13·1 answer
  • Although we talk about the velocity of an object in Uniform Circular Motion as being
    5·2 answers
  • How quickly would the 50kg box accelerate if the person applied a 570N force?
    8·1 answer
  • A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!