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3 years ago

_____cells do not contain a nucleus

Physics

1 answer:

xeze [42]3 years ago

4 0

Prokaryotic cells lack a nucleus.<span> Organisms classified as prokaryotes include bacteria and archaea. </span>

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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.

salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_{1} = u_{g}$ = 2553.6 kJ/kg

$v_{1} = v_{g} = 0.4625 m^{3}/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

$u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$\frac{Q}{m} = \Delta u$

$\frac{Q}{m} = u_{2} - u_{1}$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0

2 years ago

Mrs. Fatima has a a small sports car that she uses to drop her kids off from school that has a mass of 5000 kg. Mr. Zaid drives

sertanlavr [38]

$\textbf{Use equation of momentum}$

$p=mv$

$\text{momentum\ =\ mass x velocity}$

$\textbf{Substitute in the values}$

$5000 \times 75 = 375,000 \text{\ kg\ km/hr}$

$4200 \times 75 = 315,000 \text{\ kg\ km/hr}$

$\textbf{The\ small\ sports\ car\ will\ have\ the\ most\ momentum.}$

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2 years ago

The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is

KengaRu [80]

Answer:

Gravitational attraction is caused by the mass of an object. Since Earth is far more massive than the Moon, the gravitational force exerted on the Moon is far greater than that of the Moon on the Earth. An example of the difference: while the Moon causes tides on the Earth, the Earth has the Moon locked so that the same face (minus some wobbling) is always visible from the Earth

Explanation:

5 0

2 years ago

A 0.335 kg mass is attached to a spring and executes simple harmonic motion with a period of 0.38 s. The total energy of the sys

kramer

Answer:

$k=91.54 \frac{N}{m}$

Explanation:

The angular speed is defined as the angle traveled in one revolution over the time taken to travel it, that is, the period. Therefore, it is given by:

$\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{0.38s}\\\omega=16.53\frac{rad}{s}$

The angular frequency of the simple harmonic motion of the mass-spring system is defined as follows:

$\omega=\sqrt{\frac{k}{m}}$

Here, k is the spring's constant and m is the mass of the body attached to the spring. Solving for k:

$\omega^2=\frac{k}{m}\\k=m\omega^2\\k=0.335kg(16.53\frac{rad}{s})^2\\k=91.54 \frac{N}{m}$

6 0

3 years ago

What activity demonstrates the lowest level of intensity?

yan [13]

Since there are no choices:
aerobic training is of the lower intensity than anaerobic training. long distance running would be of lower intensity than pull ups or sprinting......

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3 years ago