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navik [9.2K]
3 years ago
14

What distinguishes a fundamental unit from a derived unit?​

Physics
1 answer:
VLD [36.1K]3 years ago
5 0

Answer:

fundamental units donot depend upon other unit of measurement but derived unit depend.

there are 7 fundamental units in si unit but there are many derived units.

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Two negative charges are both -3.0C push each other apart with a force of 19.2N. How far apart are the two charges?
Lilit [14]

06758 m

Explanation:

it is correct because that is what

5 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
Read 2 more answers
A loop with a break in it that prevents current from flowing is called a(n)_______<br> circuit.
Annette [7]

Answer:

an open circuit

Explanation:

7 0
4 years ago
The ____ phase of the SDLC includes four main activities: requirements modeling, data and process modeling, object modeling, and
nataly862011 [7]

Answer:

The system analysis phase of the SDLC includes four main activities: requirements modeling, data and process modeling, object modeling, and consideration of development strategies.

Explanation:

  • SLDC is a Software Development Life Cycle. It is basically a process of designing, developing and testing high-quality software by a software developing industries.
  • It has four processes as mentioned in the question which is a process for improving and refining the quality of software.
  • SLDC is a methodology that is used by every software company to develop a fine and high-quality software.
7 0
4 years ago
. (Use equations not the psychrometric chart) The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17oC, r
Fantom [35]

Answer:

a) The specific humidity of air is 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}.

b) The specific humidity of air is 0.464.

c) The dew-point temperature is 12.665 ºC.

Explanation:

a) The temperature of atmospheric air is considered the dry-bulb temperature, whereas the temperature of entirely saturated air is the the wet-bulb temperature. Dry bulb pressure is the atmospheric air. First we need to find the specific humidity at wet bulb temperature (\omega_{wb}), measured in kilograms of water per kilogram of dry air:

\omega_{wb} = \frac{0.622\cdot P_{wb}}{P_{db}-P_{wb}} (Eq. 1)

Where:

P_{wb} - Wet bulb pressure, measured in kilopascals.

P_{db} - Dry bulb pressure, measured in kilopascals.

Wet bulb pressure is the saturation pressure of water evaluated at wet bulb temperature, while dry bulb pressure in the pressure presented on statement. If P_{db} = 95\,kPa and P_{wb} = 1.9591\,kPa, then the specific humidity at wet bulb temperature is:

\omega_{wb} = \frac{0.622\cdot (1.9591\,kPa)}{95\,kPa-1.9591\,kPa}

\omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}

Now we use the following equation to determine the dry bulb specific humidity (\omega_{db}), measured in kilograms of water per kilogram of dry air:

\omega_{db} = \frac{c_{p,a}\cdot (T_{wb}-T_{db})+\omega_{wb}\cdot h_{fg,wb}}{h_{g,db}-h_{f,wb}} (Eq. 2)

Where:

c_{p,a} - Isobaric specific heat of air, measured in kilojoules per kilogram-Celsius.

T_{wb} - Wet-bulb temperature, measured in Celsius.

T_{db} - Dry-bulb temperature, measured in Celsius.

\omega_{wb} - Wet-bulb specific humidity, measured in kilograms of water per kilogram of dry air.

h_{fg,wb} - Wet-bulb specific enthalpy of vaporization of water, measured in kilojoules per kilogram.

h_{g,db} - Dry-bulb specific enthalpy of saturated vapor, measured in kilojoules per kilogram.

h_{f,wb} - Wet-bulb specific enthalpy of liquid vapor, measured in kilojoules per kilogram.

If we know that T_{wb} = 17\,^{\circ}C, T_{db} = 25\,^{\circ}C, c_{p,a} = 1.005\,\frac{kJ}{kg\cdot ^{\circ}C}, \omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}, h_{fg, wb} = 2460.6\,\frac{kJ}{kg}, h_{g,db} = 2546.5\,\frac{kJ}{kg} and h_{f,wb} = 71.355\,\frac{kJ}{kg}, the dry bulb specific humidity is:

\omega_{db} = \frac{\left(1.005\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (17\,^{\circ}C-25\,^{\circ}C)+\left(0.0131\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot \left(2460.6\,\frac{kJ}{kg} \right)}{2546.5\,\frac{kJ}{kg}-71.355\,\frac{kJ}{kg}  }

\omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}

The specific humidity of air is 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}.

b) Then, the relative humidity of air (\phi_{db}), dimensionless, is obtained from this expression:

\phi_{db} = \frac{\omega_{db}\cdot P_{db}}{(0.622+\omega_{db})\cdot P_{sat, db}} (Eq. 3)

Where P_{sat, db} is the saturation pressure at dry-bulb temperature, measured in kilopascals.

If we know that \omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}, P_{db} = 95\,kPa and P_{sat, db} = 3.1698\,kPa, the relative humidity of air is:

\phi_{db} = \frac{\left(9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot (95\,kPa)}{\left(0.622+9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}\right)\cdot 3.1698\,kPa}

\phi_{db} = 0.464

The specific humidity of air is 0.464.

c) The dew point temperature is the temperature at which water is condensated when air is cooled at constant pressure. That temperature is equivalent to the saturation temperature at vapor pressure (P_{v}), measured in kilopascals:

P_{v} = \phi_{db} \cdot P_{sat, db} (Eq. 4)

(\phi_{db} = 0.464, P_{sat, db} = 3.1698\,kPa)

P_{v} = 0.464\cdot (3.1698\,kPa)

P_{v} = 1.4707\,kPa

The saturation temperature at given vapor pressure is:

T_{dp} = 12.665\,^{\circ}C

The dew-point temperature is 12.665 ºC.

4 0
4 years ago
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