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kvasek [131]
2 years ago
13

at 8pm, you see that the pointer stars of the big dipper and the star polaris are arranged in a vertical line. how long, give or

take a few minutes, would you need to wait to see them arranged in a horizontal line?
Physics
1 answer:
Tju [1.3M]2 years ago
7 0

6 hours long, give or take a few minutes, you need to wait to see them arranged in a horizontal line if, at 8 pm, you see that the pointer stars of the big dipper and the star Polaris are arranged in a vertical line.

<h3 /><h3>What do you mean by pointer star?</h3>

The two prominent stars in Ursa Major known as the Pointer Stars can be utilized to locate Polaris, the North Star. The Big Dipper asterism includes the two stars Dubhe and Merak (Alpha and Beta Ursae Majoris). They trace the bowl of the Dipper's exterior. The direction of the North Star is indicated with a line that runs from Merak via Dubhe. Because it identifies the location of the celestial north pole, Polaris is significant in navigation. While other stars and constellations seem to revolve around it, the star is constantly found in the same location in the sky throughout the year.

To learn more about Polaris, Visit:

brainly.com/question/1130377

#SPJ4

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3 years ago
You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

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Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
What is the speed of a car that traveled 500 meter in 30 seconds?
GenaCL600 [577]
<h2>Greetings!</h2>

To find speed, you need to remember the formula:

Speed = distance ÷ time

So plug the given values in:

500 ÷ 30 = 16.66

<h3>So the speed is 16.66m/s (metres per second)</h3>
<h2>Hope this helps!</h2>
3 0
3 years ago
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