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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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A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r
Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

Explanation:

Given that,

Angular acceleration \alpha=\pi\ rad/s^2

Angular displacement \theta=\pi\ rad

Angular velocity \omega =2\pi\ rad/s

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}

\omega_{0}^2=\omega^2-2\alpha\theta

Where, \alpha = angular acceleration

\omega = angular velocity

Put the value into the formula

\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

\omega=\sqrt{2\pi^2}

\omega_{0}=\pi\sqrt{2}\ rad/s

Hence, The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

3 0
3 years ago
PLEASE HELP AND HURRY!!!!!!!!!
never [62]

Answer:

I believe the answer would be A. point x

4 0
4 years ago
1. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless
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Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.

4 0
3 years ago
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The strength of the force of friction depends on which two factors?
Stolb23 [73]

Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.

<h3 /><h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.

The strength of the force of friction depends on the two factors, as;

A. The objects' sizes and weights.

D. The heat generated by the friction and the types of surfaces involved.

Hence, options A and D are correct.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

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3 0
2 years ago
If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is
zysi [14]

Answer:

q_2=2.47\times 10^{-4}\ C

Explanation:

The charge on one object, q_1=9.9\times 10^{-5}\ C

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C

So, the charge on the other sphere is 2.47\times 10^{-4}\ C.

7 0
3 years ago
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