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saveliy_v [14]
3 years ago
8

A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 4.5 cm. The

two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.7 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed. (a) Calculate the angular speed of the wheel sprocket ωw, in radians per second. (b) Calculate the linear speed of the bicycle v, in meters per second, assuming the wheel does not slip across the ground. (c) If the cyclist wanted to travel at a speed of v2 = 3.5 m/s, how much time, in seconds, should elapse as the pedal makes one complete revolution?
Physics
1 answer:
ANTONII [103]3 years ago
4 0

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = <u>3.7 rad/s</u>

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

  v = (ωs)R = 7.8(65) = 507 cm/s or <u>5.1 m/s</u>

c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

ωp = 5.38(4.5 cm / 9.5 cm) = 2.55 rad/s

t = θ/ω = 2π / 2.55 = 2.463... <u>2.5 s</u>

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Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

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        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

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3 years ago
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Number of electrons, n = 6

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The measured charge of any single droplet, e=-1.6\times 10^{-19}\ C

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q=ne

n=\dfrac{q}{e}

n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}

n = 6

So, there are 6 electrons contained within the drop. Hence, this is the required solution.

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In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

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Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

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t_{A}=6.12\ sec

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s=ut+\dfrac{1}{2}at_{B}^2

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30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

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3 years ago
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     substituting values  

        \Delta  PE  =  75  *  9.8  *  540

       \Delta  PE  = 396900 \ J

3 0
3 years ago
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