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denpristay [2]
3 years ago
7

A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it

turns through 53.2 rad.
(a) How long had the wheel been turning before the start of the 7.00 s interval?
(b) What was the angular velocity of the wheel at the start of the 7.00 s interval?
Physics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

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MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

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8 0
3 years ago
Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng
klemol [59]

Answer:

Speed:

2.01x10^{8}m/s

Wavelength:

4.24x10^{-7}m

Frequency:

4.74x10^{14}Hz

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

n = \frac{c}{v} (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

v = \frac{c}{n}

v = \frac{3x10^{8}m/s}{1.490}

v = 2.01x10^{8}m/s

Hence, the speed of the laser has a value of 2.01x10^{8}m/s

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.  

To determine the frequency it can be used the following equation

c = \nu \cdot \lambda  (2)

Where c is the speed of light, \nu is the frequency and \lambda is the wavelength

Then, \nu wil be isolated from equation 2.

\nu = \frac{c}{\lambda}  (3)

Before using equation 3 it is necessary to express \lamba in units of meters.

\lambda = 632.8nm . \frac{1m}{1x10^{9}nm} ⇒ 6.328x10^{-7}m

\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}

\nu = 4.74x10^{14}s^{-1}

\nu = 4.74x10^{14}Hz

Hence, the frequency of the laser has a value of 4.74x10^{14}Hz

Wavelength:

To determine the wavelength it can be used:

v = \nu \cdot \lambda

\lambda = \frac{v}{\nu}

Where v is the speed of the laser through the polystyrene.

\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}

\lambda = 4.24x10^{-7}m

Hence, the wavelength of the laser has a value of 4.24x10^{-7}m

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Answer:

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Explanation:

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Thermometry substance is mercury or alcohol

2) Gas thermometer: This type is used to measure temperature as a result of changes in gas pressure or volume.

Thermometry substance is Gas.

3) Resistance thermometer: This type is used to measure temperature due to changes in electric resistance.

Thermometry substance is Resistance wire.

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