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zvonat [6]
3 years ago
11

A bicyclist notes that the pedal sprocket has a radius of rp = 11 cm while the wheel sprocket has a radius of rw = 4.5 cm. The t

wo sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 64 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.2 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed. Randomized Variables rp = 11 cm rw = 4.5 cm R = 64 cm t = 1.2 s show answer Correct Answer 17% Part (a) The pedal sprocket and the wheel sprocket have the same _____________________. Tangential speed at their outer edges. ✔ Correct! show answer No Attempt 17% Part (b) Calculate the angular speed of the pedal sprocket ωp, in radians per second.
Physics
1 answer:
Kay [80]3 years ago
3 0

Answer:

(a) See explanation below.

(b) ω = 5.24rad/s.

Explanation:

(a) The tangential speed at the outer edges of the sprocket and wheel are the same because the wheel does not slip and as a result the rotational kinetic energy delivered to/by the pedal transmitted uniformly throughout the chain and sprocket system. This energy causes the outer edges to move equal linear distances in equal time intervals.

Let s be the distance covered in time t

Let the tangential speed of the pedal sprocket be v1 and that of the wheel sprocket be v2

S = v1t = v2t since the time interval is the same and the wheel does not slip.

v1 = v2 = v

(b)The radius of the pedal sprocket r = 11cm = 0.11m

The pedal rotates 1 rev in every 1.2s. One revolution is equal to 2π radians.

So the angular speed is equal to ω = 2π/1.2s = 5.24rad/s

ω = 5.24rad/s

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Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit
sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

6 0
3 years ago
Describe three ways you can conserve energy in your own home
Mandarinka [93]
Turn lights off, unplug electronics, and use solar energy
4 0
3 years ago
Read 2 more answers
Potential Energy Equation (GPE = mgh)
Shkiper50 [21]

Answer:

10

Explanation:

weight=25x9.2=230

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3 years ago
A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is
hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0
3 years ago
a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second
lawyer [7]

Answer:

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b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

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3 years ago
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