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butalik [34]
4 years ago
10

viewed from earth two stars form an angle of 76.04 degrees. StarA is 23.30 light years from earth star Bis 34.76 light years fro

m earth sketch a diagram modeling this situation and find how many light years the stars are from eachother ...?
Physics
1 answer:
kirill115 [55]4 years ago
4 0
The observation point on Earth and the two stars form a triangle. The two sides of the triangle are 23.3 ly and 34.76 ly and their included angle is 76.04°. We can use the cos rule to find the third side, which is the distance between the two stars.
c² = a² + b² - 2abCos(C)
c² = (23.3)² + (34.76)² - 2(23.3)(34.76)Cos(76.04)
c = 36.88 light years.
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To prevent dehydration when exercising on hot and humid days it is vital to:
Nostrana [21]
Drink water before, during, and after your exercise activity! i would rlly appreciate brainliest:D
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3 years ago
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At which point is there the most potential energy? At which point is there the most kinetic energy?
AleksandrR [38]

Answer:

The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy

5 0
3 years ago
Doug, who runs track for his high school, was challenged to a race by his younger brother, matt. Matt started running first, and
eduard

The speed of Matt is 10 mph.

Doug runs 2 miles an hour faster than Matt, so let Matt’s speed equal x miles per hour. Then Doug’s speed equals x + 2 miles per hour. Each lap is one-quarter of a mile, so Doug runs 1.5 miles in the time it takes Matt to run 1.25 miles.

Rate of Matt is x

Rate of Dough is (x + 2)

Time taken by Matt is 1.25/x

Time taken by Dough is 1.25/(x + 2)

Distance covered by Matt is 1.25

Distance covered by Dough is 1.5

Dough and  Matt  took the same amount of time from the time Doug started, so make an equation by setting the two times in the chart equal to each other, and then solve for x:

            \frac{1.5}{(x + 2)} = \frac{1.25}{x}

              1.5x = 1.25(x + 2)

              1.5x = 1.25x + 2.5

           0.25x = 2.5

                   x = 10

So Matt ran at 10 miles per hour.

To know more about time, speed and distance, visit: brainly.com/question/26046491

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7 0
1 year ago
Three liquids are at temperatures of 4 ◦C, 24◦C, and 29◦C, respectively. Equal masses of the first two liquids are mixed, and th
Scilla [17]

Answer:

T₁₃ = 24.1°C

Explanation:

Given

m = mass of each liquids (all masses are equal)

C₁  = specific heat of the first liquid

C₂ = specific heat of the second  liquid

C₃ = specific heat of the third liquid

T₁ = 4°C (temperature of the first liquid)

T₂ = 24°C  (temperature of the second liquid)

T₃ = 29°C  (temperature of the third liquid)

​Temperature of 1+2 liquids mix: T₁₂ = 21°C

​Temperature of 2+3 liquids mix: T₂₃ = 26.1°C  

Temperature of 1+3 liquids mix: T₁₃ = ?

We can apply the relation ∑Q = 0

We assume the system is isolated and the process is adiabatic.

<u>Mix 1</u>:

Q₁ + Q₂ = 0

where

Q₁ = m*C₁*(T₁-T₁₂)

and

Q₂ = m*C₂*(T₂-T₁₂)

then

m*C₁*(T₁-T₁₂) + m*C₂*(T₂-T₁₂) = 0

⇒ C₁*(T₁-T₁₂) + C₂*(T₂-T₁₂) = 0

⇒ (4°C-21°C)*C₁ + (24°C-21°C)*C₂ = 0

⇒ -17°C*C₁ + 3°C*C₂ = 0

⇒ C₁ = (3/17)*C₂ = 0.176*C₂     (i)

<u>Mix 2</u>:

Q₂ + Q₃ = 0

where

Q₂ = m*C₂*(T₂-T₂₃)

and

Q₃ = m*C₃*(T₃-T₂₃)

then

m*C₂*(T₂-T₂₃) + m*C₃*(T₃-T₂₃) = 0

⇒ C₂*(T₂-T₂₃) + C₃*(T₃-T₂₃) = 0

⇒ (24°C-26.1°C)*C₂  + (29°C-26.1°C)*C₃ = 0

⇒ -2.1°C*C₂ + 2.9°C*C₃ = 0

⇒ C₃ = 0.724*C₂      (ii)

<u>Mix 3</u>:

Q₁ + Q₃ = 0

where

Q₁ = m*C₁*(T₁-T₁₃)

and

Q₃ = m*C₃*(T₃-T₁₃)

then

m*C₁*(T₁-T₁₃) + m*C₃*(T₃-T₁₃) = 0

⇒ C₁*(T₁-T₁₃) + C₃*(T₃-T₁₃) = 0

⇒ (4°C-T₁₃)*C₁ + (29°C-T₁₃)*C₃ = 0

If we use the following relations  C₁ = 0.176*C₂ and C₃ = 0.724*C₂  we obtain

(4°C-T₁₃)*0.176*C₂ + (29°C-T₁₃)*0.724*C₂ = 0

⇒ (4°C-T₁₃)*0.176 + (29°C-T₁₃)*0.724 = 0

⇒ 0.706°C - 0.176*T₁₃ + 21°C - 0.724*T₁₃ = 0

⇒ 0.9*T₁₃ = 21.706°C

⇒ T₁₃ = 24.1°C

5 0
3 years ago
Ask Your Teacher The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s
adell [148]

Answer:

\theta = 20\ rev

Explanation:

Case 1

Given,

initial angular speed = 0 rev/s

final angular speed = 2 rev/s

time, t = 7 s

angular acceleration of the washer

\alpha = \dfrac{\omega_f-\omega_i}{t}

\alpha = \dfrac{2-0}{7}

\alpha = 0.286\ rev/s^2

using equation of rotational motion

\omega_f^2 = \omega_i^2 + 2 \alpha \theta_1

2^2 = 0^2 + 2\times 0.286 \times \theta_1

\theta_1 = 7 rev

Case 2

Given,

initial angular speed = 2 rev/s

final angular speed = 0 rev/s

time, t = 12 s

angular acceleration of the washer

\alpha = \dfrac{\omega_f-\omega_i}{t}

\alpha = \dfrac{0-2}{13}

\alpha = -0.154\ rev/s^2

using equation of rotational motion

\omega_f^2 = \omega_i^2 + 2 \alpha \theta_2

0^2 = 2^2 + 2\times (-0.154) \times \theta_2

\theta_2 = 13 rev

total revolution in this case

\theta = \theta_1 + \theta_2

\theta =7 +13

\theta = 20\ rev

total revolution of the washer is equal to 20 rev.

6 0
3 years ago
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