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tester [92]
3 years ago
8

which option below best explains why the second law of thermodynamics is not violated when heat flies from a cold freezer and in

to the much warmer room?
Physics
1 answer:
Rainbow [258]3 years ago
8 0

Explanation:

  1. i am very sorry but you didn't give any options so i can't help you

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Help it’s multiple choice 11 through 15 please!
riadik2000 [5.3K]

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

u_{k} = 0.4

• Force of friction =

u_{k} \times  \: normal \:  \:  \: force \\  = 0.4 \times 20N \\  = 8N

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• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

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