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Andrei [34K]
3 years ago
11

Brainliest!! Please help with number 4 and please give me the equation too!! Brainliest!!!

Physics
2 answers:
Thepotemich [5.8K]3 years ago
8 0

Answer:

Oooo someone is writing a answer. (Also im new to this so Idk what to really do.)

Explanation:

viva [34]3 years ago
4 0

Answer:

3.9×10⁸ m

Explanation:

If x is the distance between the Earth and the Moon, the the light travels a distance of 2x in 2.6 seconds.

Distance = rate × time

2x = (3×10⁸ m/s) (2.6 s)

x = 3.9×10⁸ m

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I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second
Degger [83]

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

3 0
4 years ago
Read 2 more answers
. (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
The lens in the eyepiece of a telescope magnifies light by bending it. What is this bending called?
tresset_1 [31]
I believe that it is called refracting...
7 0
3 years ago
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What type of shoes are worn by the mountaineers and why​
Ilia_Sergeevich [38]
Mountaineers wear shoes called approach shoes or rock shoes. They are spiky on the rough. They wear them to increase the action of friction on the ground.
5 0
3 years ago
If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the induc
mel-nik [20]

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

5 0
4 years ago
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