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Brrunno [24]
3 years ago
10

Triangulation is the preferred method of calculation for all astronomical distances.

Physics
2 answers:
yan [13]3 years ago
8 0
<span>
if it was possible to triangulate all objects it would give the most accurate answer.  So yes, it is the prefered method.  However as this is not the case other methods are used, less accurate but more practical</span>
never [62]3 years ago
4 0
The answer would be : True.

Hope this helps !

Photon
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A car travels in a straight line from a position 50 m to your right to a position 210 m to your right in 5 sec. a. What is the a
Verizon [17]

Answer:

a) The average velocity is 32 m/s

b) See the attached figure. Slope of the line = 32.

Graphic: a line that passes through the points (0; 50) and (5; 210)

Explanation:

a) The average velocity is calculated as the displacement over time:

v = ΔX / Δt

where

ΔX : displacement ( final position - initial position)

Δt : time (final time - initial time)

Considering the origin of the reference system as the position where the observer is:

ΔX = 210 m - 50 m = 160 m

Δt  = 5 s

v = 160 m / 5 s = 32 m/s

b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:

slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.

4 0
3 years ago
How long will it take an object to hit the ground if it is dropped from a hight of 176.4 meters ​
Oliga [24]

There's a short handy formula for that.

If the object is just dropped and not tossed, and it's not affected by air resistance on the way down, then the distance it falls in T seconds is

D = (1/2) (gravity) (T²)

For this problem . . .

176.4 m = (1/2) (9.8 m/s²) (T²)

Divide each side by  (4.9 m/s²) :

T² = (176.4 m) / (4.9 m/s²)

T² = (36 s²)

Take the square root of each side:

<em>T = 6 seconds</em>

5 0
3 years ago
4. How much milk at 5° C needs to be added to 250 g of coffee at 90° C to make the coffee drinkable at 60° C?
kenny6666 [7]

Answer:

dino :)

Explanation:

7 0
3 years ago
The diagram below shows the movement of matter in a portion of the water cycle.
Alisiya [41]

Answer:

d) precipitation

Hope it helps you

And if you want to, pls mark it as the brainliest answer

7 0
3 years ago
Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volu
timurjin [86]

Answer:

1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2) W₁ = 0 J

3) P = 41.66 x 10³ Pa = 41.66 KPa

4) v = 1618.72 m/s

5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

6) W₂ = - 7.33 KJ

Explanation:

1)

The heat transfer for a constant volume process is given by the formula:

ΔQ₁ = ΔU = n Cv ΔT

where,

ΔQ₁ = Heat transfer during constant volume process

ΔU = Change in internal energy of gas

n = No. of moles = 4.4 mol

Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k

Therefore,

ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)

<u>ΔQ₁ = 10.97 x 10³ J = 10.97 KJ</u>

<u></u>

2)

Since, work done by gas is given as:

W₁ = PΔV

where,

ΔV = 0, due to constant volume

Therefore,

<u>W₁ = 0 J</u>

<u></u>

4)

The average kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

but, K.E is also given by:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

mv² = 3KT

v = √(3KT/m)

where,

v = average speed of gas molecue = ?

K = Boltzman Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 500 K

m = mass of a molecule = 7.9 x 10⁻²⁷ kg

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]

<u>v = 1618.72 m/s</u>

<u></u>

3)

From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²

where,

P = pressure of gas = ?

m = Mass of Gas = (Atomic Mass)(No. of Atoms)

m = (Atomic Mass)(Avogadro's Number)(No. of Moles)

m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)

m = 0.021 kg

ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³

Therefore,

P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²

<u>P = 41.66 x 10³ Pa = 41.66 KPa</u>

<u></u>

5)

The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ =  n Cp ΔT

where,

ΔQ₂ = Heat transfer during constant pressure process

n = No. of moles = 4.4 mol

Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k

Therefore,

ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)

<u>ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ</u>

<u>Negative sign shows heat flows from system to surrounding.</u>

<u></u>

6)

From Charles' Law, we know that:

V₁/T₁ = V₂/T₂

V₂ = (V₁)(T₂)/(T₁)

where,

V₁ = 0.44 m³

V₂ = ?

T₁ = 500 K

T₂ = 300 k

Therefore,

V₂ = (0.44 m³)(300 k)/(500 k)

V₂ = 0.264 m³

Therefore,

ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³

Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)

<u>W₂ = - 7.33 KJ</u>

<u>Negative sign shows that the work is done by the gas</u>

4 0
3 years ago
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