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damaskus [11]
3 years ago
5

A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,

it will have turned through approximately how many revolutions?
1. 65
2. 39
3. 52
4. 26
5. 13
Physics
1 answer:
ZanzabumX [31]3 years ago
8 0

Answer:

5) 13 revolutions (approximately)

Explanation:

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2α*θ Formula (1)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

Data:

ω₀ = 18 rad/s

ωf = 0

α = -2 rad/s²  ; (-) indicates that the wheel is slowing

Revolutions calculation that turns the wheel until it stops

We apply the formula (1)

ωf²= ω₀² + 2α*θ

0 = (18)² + 2( -2)*θ

4*θ =  (18)²

θ =  (18)²/4 = 81 rad

1 revolution = 2π rad

θ = 81 rad * 1 revolution / 2πrad

θ =  13 revolutions approximately

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If a marathon runner averages 9.51 mi/hr, how long does it take him to run a 26.220-mile marathon. Express your answers in hours
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Explanation:

Speed of the marathon runner, v = 9.51 mi/hr

Distance covered by the runner, d = 26.220 mile

Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :

v=\dfrac{d}{t}

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

                           w=\frac{2*wavelength*d}{a}

where, d is the distance between slit and the screen, and a is the slit width.

  • Thus, distance between slit and the screen is,

                           d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

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The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
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Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

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