2 years ago

Which law states that absolute zero cannot be reached?

2 answers:

5 0

The third law of thermodynamics,the principle of temperature.

This law states that the entropy at 0 is always equel to 0.
This means that it is impossible to cool down a perfect 0 or absolute 0(-273.15 C)

nalin [4]2 years ago

4 0

The answer is third law

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2 years ago

PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!

pav-90 [236]

Answer:

50 N

Explanation:

Let the natural length of the spring = L

100 = k(40 - L) (1)

200 = k(60 - L) (2)

(2)/(1): 2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

3 0

1 year ago

Question 3 Please, i need help! Thank you

Tom [10]

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4 0

3 years ago

WILL MARK BRAINLIST

ryzh [129]

Answer:

Definition of envelop

transitive verb

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3 0

1 year ago

A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

6 0

3 years ago