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Marrrta [24]
3 years ago
10

A rotating wheel requires a time Δt = [01]_____________________ to rotate 37.0 revolutions. Its angular speed at the end of the

interval is ω = 75.9 rad/s. What is the constant angular acceleration (rad/s2 ) of the wheel
Physics
1 answer:
Sidana [21]3 years ago
8 0

Answer:

Explanation:

Initial angular velocity ω₁ = 0 , final angular velocity ω₂ = 75.9 rad /s

angle rotated =  θ

= 37 x 2π

= 74 π

The formula for angular velocity

ω₂² =  ω₁² + 2αθ , α is angular acceleration

75.9² = 0 + 2 α x 74 π

α = 75.9² / 2  x 74 π

= 12.396 rad / s²

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Light travels in a transparent material at 2.5 x 10 m/s. Find the index of refraction of the
Andreyy89
I would say 5.6 to the 4 and I’ll I did was add it to both side
3 0
3 years ago
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
34.6 cL= (blank) hL convert
shepuryov [24]

<u>Answer</u>

0.00346 hL

<u>Explanation</u>

cL means Centilitre while hL means Hectolitre.

10,000 cL = 1 hL

∴ 34.6 cL = 34.6/10,000  hL

                = <em>0.00346 hL</em>

3 0
4 years ago
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
What is your volume of the object?<br> 30 cm3<br> 35 cm3<br> 42 cm3<br> 54 cm3
Alinara [238K]
The answer is 54 cm3
3 0
3 years ago
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