Please help me with this question...

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.

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3 years ago

Ball 1, with a mass of 100 g and traveling at 10 m/s, collides head-on with ball 2, which has a mass of 300 g and is initially a

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There is 4000 balls in the earth of the world why is that so hard for you

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2 years ago

A skateboarder rolls horizontally off the top of a staircase and lands at the bottom. The staircase has a horizontal length of 1

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Answer: 18.8 m/s

Explanation:

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3 years ago

A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and pa

VLD [36.1K]

The ball takes 0.2 sec to travel the height of the window, which is 2 m.

Then the speed of the ball at that time was

$v = \frac{d}{t}$

$v = \frac{2}{0.20}$

$v = 10\ \frac{m}{s}$

We know that:

$v = v_0 -gt$

Where $v_0$ is the initial velocity.

So:

$10 = v_0 -9.8(2)$

$10 + 9.8(2) = v_0$

$v_0 = 29.6\ m / s$

Then we have the equation for the position as a function of time.

$r = r_0 + v_0t - \frac{1}{2}gt^2$

Where

$r_0$ = initial position

r = position as a function of time

$v_0$ = initial velocity

g = acceleration of gravity

t = time.

If the ball is thrown from the ground then:

$r_0$ = 0 m

We want to find now the distance between the window and the ground.

When the ball reaches the bottom of the window t = 1.8s

So:

$r = 0 + 29.6(1.8) - 0.5(9.8)(1.8)^2$

$r = 37,404$ m

The window is 37,404 m high

Finally, to know how high the ball rises we must know at what moment the vertical velocity of the ball is zero.

$v = v_0 -gt\\\\0 = v_0 -gt\\\\gt = v_0\\\\t = \frac{29.6}{9.8}$

$t = 3.02\ s$

Now we replace t in the position equation

$r = 0 + 29.6(3.02) -0.5(9.8)(3.02)^2$

$r = 44.70\ m$

The ball reaches up to 44.70 m in height.

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3 years ago

The binding of acetylcholine to its receptor at the neuromuscular junction causes the opening of a

Rufina [12.5K]

Answer: opening of the nicotinic acetylcholine receptor channels.

Explanation:

Neuromuscular junction is a special junction formed between a motor neurone and a muscle fibre. The junction is fortified with nerves and receptors that helps in the transmission of signals from the motor neurone to the muscle fibre in order to bring about the desired voluntary movements through muscular contraction.

Nicotinic acetylcholine receptor are activated through the binding of acetylcholine at the neuromuscular junction. This action leads to influx of sodium ions to accomplish endplate potential.

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3 years ago