All categories

3 years ago

What do we call the point where molecules would have 0 kinetic energy?

Physics

1 answer:

kogti [31]3 years ago

7 0

Absolute Zero is the name for when molecules have 0 kinetic energy.

You might be interested in

You wanted to use your electric grill in your backyard. You need an extension cord. What will be the requirement of your extensi

Troyanec [42]

Answer:

you need to be able to have long enough to reach and have it far away from things that are going to cause accidents

8 0

3 years ago

Read 2 more answers

To escape from a horrible fire, two people are forced to jump from the third story of a burning building onto solid concrete. Wh

erica [24]

The bouncing person because the bounce helped him survive

8 0

3 years ago

Someone who snorts and stop breathing during sleep may be suffering from: a. Catoplexy b. Narcolepsy c. Sleep Apnea d. Insomnia

klio [65]

Answer:

C. Sleep Apnea

Explanation:

Sleep Apnea is when someone who snores and stops breathing while sleeping

6 0

2 years ago

The desert sand is very hot during the day and very cold at night. What does this tell you about its specific heat capacity?

erma4kov [3.2K]

Answer: The specific heat capacity is very low.

Explanation:

The specific heat capacity of a body is defined as the heat energy required by a body to cause a unit change in its temperature. The value is over low that is why it is easier for the desert sand to easily get very hot during the day. Conversely, it is very easy for the desert sand to lose it's heat a cool breeze pass over it in the night making it very cold in the night. This value also defines how long the desert sand can retain heat. Therefore, the desert sand has a low specific heat capacity.

8 0

3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

1 year ago