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mars1129 [50]
3 years ago
9

A star is located 12.5 parsecs away from us. What is its parallax angle? D = 1/p

Physics
2 answers:
mars1129 [50]3 years ago
8 0

Answer:

<h2>The parallax angle is 0.08 seconds of arc.</h2>

Explanation:

In the image attached you can observe where is the parallax angle placed.

Now, this parallax angle is inversely proportional to the distance measured in parsecs. That means, the greater the distance, the smaller the parallax angle.

Mathematically, it's defined as

D=\frac{1}{p}

Where D=12.5 pc, replacing this value, we have

D=\frac{1}{p}\\p=\frac{1}{D}=\frac{1}{12.5 pc}  =0.08''

Therefore, the parallax angle is 0.08 seconds of arc.

Rudiy273 years ago
7 0
You are given a star that is located 12.5 parsecs away from us (let us assume that the reference point is on earth). You are tasked to find the parallax angle. The equation to get the parallax angle is D = 2 (AU)/ tanФ wherein D is the distance of the star from the earth, AU is the distance of earth from the sun and Ф is the parallax angle. Note that 1 parsec = 1,000,000,000,000 million meters and that the distance of earth to the sun is about 149,600,000,000 meters.

D = 2 (AU)/ tanФ
tan Ф = 2 (AU)/D
Ф = tan⁻¹ [ 2(AU) / D]
Ф = tan⁻¹ [ 2(149,600,000,000) / 12,500,000,000,000]
Ф = 1.37°
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If the total resistance of an air conditioner is 25.4 Ω and it is using energy at a rate of 2083 W, on what voltage is it operat
strojnjashka [21]
One of the formulas for calculating electrical power is

Power = (voltage)² / (resistance)

Since this problem wants us to find the voltage, let's
re-arrange the formula.
Multiply each side by (resistance):

(Voltage)² = (power) x (resistance)

Take the square root of each side:

Voltage = √ (power x resistance)

THERE's the formula we can use to find the voltage for this problem.

Voltage = √ (2,083 W x 25.4 Ω)

Voltage = √ (52,908.2 W-Ω )

Voltage = 230 volts
6 0
4 years ago
A force of 70 N is inclined at an angle
Bad White [126]

Answer:

The horizontal component of the 70 N force = 35·√3 N

The vertical component of the 70 N force = 35 N

Explanation:

The magnitude of the given force, F = 70 N

The angle of inclination of the given force to the horizontal, θ = 30°

By the resolution of forces, we can resolve the given force, F, into its horizontal component, Fₓ, and vertical components, F_y, as follows;

The horizontal component of the given force = Fₓ = F × cos(θ)

Substituting the values, we have;

The horizontal component of the 70 N force, Fₓ = 70 × cos(30°) = 35·√3 N

The vertical component of the given force = F_y = F × sin(θ)

Substituting the values, we have;

The vertical component of the 70 N force, F_y = 70 × sin(30°) = 35 N.

5 0
3 years ago
A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?
kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

v=\frac{S}{t}

Rearranging the equation, we can write

S=vt

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m


2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s

5 0
4 years ago
20 J of energy was used to
Leto [7]

Answer:

0.3m

Explanation:

Given parameters;

Elastic energy  = 20J

Spring constant = 445N/m

Unknown

Final extension of the spring  = ?

Solution:

The elastic potential energy of a stretched spring can be determined using the expression below;

         

           EP  = \frac{1}{2} k e²

k is the spring constant

e is the extension

  Insert the parameters and solve;

          20  = \frac{1}{2} x 445 x e²

        multiply those sides by 2;

         2(20) = 445e²

            40  = 445e²

              e² = \frac{40}{445}   = 0.09

              e  = √0.09 = 0.3m

6 0
3 years ago
Place the charge q= 2.10-5c at the origin o of the oxy-axis system. consider two points a(6;0) and b(3;4) on the axes with unit
laila [671]

The work done by the electric field when moving the charge from a to b is 0.0378 J.

<h3>Distance between a and b</h3>

r = √[(a₂ - a₁)² + (b₂ - b₁)²]

r = √[(3 - 6)² + (4 - 0)²]

r = 5 m

<h3>Work done in moving the charge</h3>

W = Fr

W = kq₁q₂/r

W = (9 x 10⁹ x 2.1 x 10⁻⁵ x 1 x 10⁻⁶)/(5)

W = 0.0378 J

Thus, the work done by the electric field when moving the charge from a to b is 0.0378 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

4 0
2 years ago
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