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andrew11 [14]
3 years ago
14

2.5

Physics
1 answer:
Anit [1.1K]3 years ago
5 0

Answer:

what is this I don't understand the question

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A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t
White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

        v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

         w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

         w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

        \frac{v_o}{r_o} = \frac{v_1}{r_1}

        v1 = \frac{r_1}{r_o} \ \ v_o

let's calculate

       v₁ = \frac{1.50}{1.00} \ \ 25.0

       v₁ = 37.5 cm / s

4 0
3 years ago
Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r
bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

where,

T = Radiated surface temperature

\lambda_{max} = peak wavelength

(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

<u>T = 2987.6 k</u>

(b)

here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}

<u>T = 19986.2 k</u>

6 0
3 years ago
A ball is launched from ground level at 20 m/s at an angle of 40° above the
DedPeter [7]

(a) The ball's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :

0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²

0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )

<em>t</em> = 0   or   (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 <em>g t</em>

<em>t</em> = (40 m/s) sin(40º) / <em>g</em>

<em>t</em> ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So

0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>

where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :

<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)

<em>y</em> ≈ 8.4 m

8 0
3 years ago
Does the atmosphere contain 21% nitrogen in 78% oxygen
Alex777 [14]

Answer:

An atmosphere is the layers of gases surrounding a planet or other celestial body. Earth's atmosphere is composed of about 78% nitrogen, 21% oxygen, and one percent other gases

7 0
3 years ago
For work to be done on an object,
djverab [1.8K]

Answer:B

Explanation:

For work to be done, the object must move some distance as a result of a force

3 0
3 years ago
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