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3 years ago

Every force has one and only one 3rd law pair force. every force has one and only one 3rd law pair force.

Physics

2 answers:

riadik2000 [5.3K]3 years ago

8 0

The correct answer is a. It is true

Likurg_2 [28]3 years ago

7 0

I would have to say B failed because I think I read something about it being only 2law not 3

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A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

3 0

3 years ago

In a race, a runner traveled 12 meters in

Nastasia [14]

Since it is a race, the runner start from rest, so we can use the Velocity Hourly Equation to describle yours moviment.

$S=S_{o}+V_{o}t+ \frac{at^2}{2} \\ S=\frac{at^2}{2}$

Entering the unknowns, we have:

$S=\frac{at^2}{2} \\ 12*2=a*4^2 \\ \boxed {a=1.5m/s^2}$

Number 2

If you notice any mistake in my english, please let me know, because i am not native.

3 0

3 years ago

Which of the following represents a covalent compound? KF NaCl SBr3 Mg2O

slava [35]

Answer:

SBr3 is a covalent compound

5 0

4 years ago

A uniform disk of 10 kg and radius 4.0 m can rotate in a horizontal plane about a vertical axis through its center. The disk is

o-na [289]

Answer:

18.75 rad/s

Explanation:

Moment of inertia of the disk;

I_d = ½ × m_disk × r²

I_d = ½ × 10 × 4²

I_d = 80 kg.m²

I_package = m_pack × r²

Now,it's at 2m from the centre, thus;

I_package = 5 × 2²

I_package = 20 Kg.m²

From conservation of momentum;

(I_disk + I_package)ω1 = I_disk × ω2

Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.

Thus;

Plugging in the relevant values, we obtain;

(80 + 20)15 = 80 × ω2

1500 = 80ω2

ω2 = 1500/80

ω2 = 18.75 rad/s

5 0

3 years ago

Which of these could NOT be used to find the mechanical advantage of an inclined plane?

34kurt

The "c) percent efficiency" could not be used to find the mechanical advantage of an inclined plane. There are two formulae that could be used to determine the mechanical advantage of an inclined plane which stated as MA = Length/rise and Wout=Win. MA is the mechanical advantage, Wout is the output force, Win is the input force, and "rise" is the height of the inclined plane.

8 0

3 years ago