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3 years ago

Please help need to know ASAP.

2 answers:

8 0

Answer: because she is pedalling at her maximum speed produced by the maximum force applied. At constant speed, acceleration is equal to zero.

Explanation:

Pedalling of bicycle involves application of force. The force applied produces circular motion to the tires which eventually transform into linear speed.

V = wr

Where V = linear speed

W = angular speed

r = radius.

Change in speed V will lead to acceleration or deceleration depending on increase or decrease in speed.

If she stops accelerating, then, she must have applied force that makes her pedalling at maximum speed. She is also maintaining this uniform (constant ) speed. After reaching her maximum speed.

At constant speed, acceleration = 0

Base on this explanation, even though she is still pedalling as fast as she can, which at constant speed, she will stop accelerating and her speed reaches a maximum value because she is pedalling at her applied maximum force.

emmainna [20.7K]3 years ago

3 0

Answer:

Resistive forces get bigger as the speed increases and eventually they reach the point where the pedalling forces become balanced and therefore there is no further acceleration.

Explanation:

Hope this helps:))

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All interactions between the atmosphere and the geosphere involve

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Answer:

The geosphere consists of the solid Earth and the atmosphere consists of the gaseous components in the air. Thus, the answer is C.

Explanation:

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3 years ago

What is the change in velocity of a 22-kg object that experiences a force of 15 N for

vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

${ \tt{15 = (22 \times a)}} \\ { \tt{a = \frac{15}{22} \: {ms}^{ - 2} }}$

From first Newton's equation of motion:

${ \bf{v = u + at}}$

Change = v - u:

${ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \: {ms}^{ - 2} }}$

3 0

3 years ago

A copper (Young's modulus 1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to en

Nina [5.8K]

We have that for the Question, it can be said that the amount by which the length of the stack decreases is

$dl'=3.621*10^{-4}m$

From the question we are told

A copper (Young's modulus 1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to end, as in the drawing. Each cylinder has a radius of 0.24 cm.

A compressive force of F = 7900 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.

Generally the equation for copper cylinder is mathematically given as

$dl=\frac{Flo}{yA}$

$dl=\frac{7900*3*10^-^2}{1.1*10^{11}*\pi(0.24*10^{-2})^2}$

$dl=1.19064778*10^-^4$

Generally the equation for brass cylinder is mathematically given as

$dl=\frac{7900*5*10^-^2}{9*10^{10}*\pi(0.24*10^{-2})^2}$

$dl=2.43*10^{-4}$

Therefore Total change in length

$dl'=1.191*10^-^4+(2.43*10^{-4})$

$dl'=3.621*10^{-4}m$

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brainly.com/question/23379286

3 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

3 years ago

A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinde

Romashka [77]

Answer:

(a)10.5 rad/s2

(b) 20.9 rev

Explanation:

As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law

T = mg = 53*9.81 = 519.93 N

Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is

To = TR = 519.93 * 0.36 = 187.17 Nm

This solid cylinder would have a moment of inertia around it's rotating axis of:

$I = \frac{mR^2}{2} = \frac{275 * 0.36^2}{2} = 17.82kgm^2$

(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder

$\alpha = \frac{To}{I} = \frac{187.17}{17.82} = 10.5 rad/s^2$

(b) With such constant angular acceleration, the angle it would make after 5s is

$\theta = \frac{\alphat^2}{2} = \frac{10.5*5^2}{2} = 131.3 rad$

Since each revolution equals to $2\pi rad$ of angle, we can calculate the number of revolution it makes

$\frac{\theta}{2\pi} = \frac{131.3}{6.28} \approx 20.9 rev$

(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad

$\theta R = 131.3*0.36 = 47.27 m$

3 0

3 years ago