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3 years ago

Why can't there be a number lower than absolute zero

2 answers:

5 0

Absolute zero is not about numbers. It's about temperature, and the
motion of molecules in gases.

You know that the temperature we feel with our skin is the result of the
average speed of all the tiny molecules zipping around or vibrating in
the solid, liquid, or gas.

The faster they're all moving, the warmer the substance feels to us.
The slower they're all moving, the cooler the substance feels to us.

When molecules slow down to zero and lose all of their kinetic energy,
that temperature is what we call 'absolute zero' ... if they're not moving
at all, then they can't move any slower.

Dafna1 [17]3 years ago

5 0

<span>Temperature lower than absolute zero is not possible. Lower than absolute zero a gas contain negative pressure and volume which is not possible. temperature is the energy created by moving atoms absoulte zero is a complete stand still you can make particles move to the end of the universe but the coldest temperature will be stuck at the point where atoms are motionless</span>

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You are driving at 25 m/s with your cruise control on when you see a fallen tree in the road. It takes you 0.30 s to put on the

xxTIMURxx [149]

Explanation:

Given data

velocity v= 25m/s

The time it takes to put on brake t= 0.3s

the distance covered when the brake was put on is

v=s/t

s= v*t

s= 25*0.3s

s= 7.5m

hence the distance covered is 7.5m

Also the rate of decrease in aceleration is 5m/s^2

we can also calculate the distance covered at this rate

v^2=u^2+2as

25^2= 0+2*5*s

625=10s

divide both sides by 10

s=625/10

s= 62.5m

The total distance covered between putting on the brakes and decelareation is 7.5+62.5= 70m

Given that the tree is 75m ahead, the car would not hit the tree

3 0

3 years ago

Raul wants to increase his upper body and cardiovascular strength. He plans to go to the gym and use his backyard swimming pool

Mkey [24]

Answer: An exercise, activity, or workout that targets a specific training need Progression - Gradually increasing your skill level and ability over time Overload -Working your body harder than usual to improve how it functions Reversibility - Losing what you have gained by taking time off Tedium - getting bored by doing the same thing

Explanation:

4 0

2 years ago

Your outlets at home are rated at 120 V, i.e. the two prongs have on average a potential difference of 120V. If you transfer 2.7

hodyreva [135]

Answer:

E =230.4 MJ

Explanation:

As 1 mole of electron = 6X 10^23 particles.

charge of an electron is 1.6 X 10 ^-19 C

Finding Charge:

(6X10^23 ) (2.7)(1.6X10^-19 C)

i.e. 192 K C

now to find the energy released from electrons

V=E/q

E=V X q

i.e E = 120 V X 192 K C

E =230.4 MJ

4 0

3 years ago

Which of the following is NOT a level of home monitoring?

Bingel [31]

<h2>Answer:</h2>

d. electronic monitoring

<h2>Explanation:</h2>

Home monitoring in its simplest term is the act of restricting the movement of an entity, object or an individual to a specified location, usually their homes.

In criminology, there are various levels of home monitoring. Some of them are;

i. <em>Curfew</em>: This type of home monitoring requires that the persons being monitored be indoors/at home at specified hours.

ii. <em>Home incarceration</em>: This level of home monitoring requires that the persons being monitored be indoors/at home at all times except in times of medical urgencies.

iii. <em>Home detention</em>: This seems to be one of the less strict forms of home monitoring. It requires that the persons being monitored be indoors/at home at all times except when necessary. For example, they could go to school, their workplace, to the market and so on.

Electronic monitoring is not a form of home monitoring as it allows the person being monitored to move freely. They might just contacted via telephone or any other means, at intervals to know their whereabouts.

5 0

2 years ago

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

2 years ago