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Andrews [41]
3 years ago
8

Two ladybugs sit on a rotating disk that is slowing down at a constant rate. The ladybugs are at rest with respect to the surfac

e of the disk and do not slip. Ladybug 1 is halfway between ladybug 2 and the axis of rotation. How does the rotational speed of ladybug 1 compare to ladybug 2?
Physics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

The two ladybugs have same rotational (angular) speed

Explanation:

The rotational (angular) speed of an object in circular motion is defined as:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval considered

Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle \theta in the same time t: this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.

On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:

v=\omega r

where r is the distance from the axis: and since the two ladybugs are located at different r, they have different linear speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v
mihalych1998 [28]

Answer:

a.

\displaystyle a(0 )=8.133\ m/s^2

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=0.52\ m/s^2

b.\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. t=9.9 \ sec

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

\displaystyle V(t)=a(1-e^{bt})

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

\displaystyle a=11.81\ ,\ b=-0.6887

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

\displaystyle V(t)=11.81(1-e^{-0.6887t})

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})

\displaystyle a(t)=8.133547\ e^{-0.6887t}

For t=0

\displaystyle a(0)=8.133547\ e^o

\displaystyle a(0 )=8.133\ m/s^2

For t=2

\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}

\displaystyle a(4)=0.52\ m/s^2

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C

\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

\displaystyle x(0)=0=>11.81\times1.45201+C=0

Solving for C

\displaystyle c=-17.1482\approx -17.15

Now we complete the equation for the distance

\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100

Rearranging

\displaystyle t+1.45\ e^{-0.6887t}=9.92

We define an auxiliary function f(t) to help us find the value of t.

\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92

Let's try for t=9 sec

\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92

Now with t=9.9 sec

\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184

That was a real close guess. One more to be sure for t=10 sec

\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).  

At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}

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3 years ago
If a participant were holding two different weights in their hands and the jnd for a 10-gram weight was 1 gram, what should the
Nataliya [291]

The jnd for a 100-gram weight, according to Weber's law will be 10 gram.

<h3>What is Weber's law?</h3>

It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.

In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;

= 100 / 10

= 10 gram

Therefore, jnd for a 100-gram weight, according to Weber's law will be 10 grams.

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1 year ago
Energy input remains constant and voltage remains the same in a circuit, but current decreases. Which must be happening?
RUDIKE [14]
A) the resistance is increasing

Hope this helped!
6 0
4 years ago
Electrical power is transmitted from power plants to consumers–sometimes over very long distances– through conducting power line
sergeinik [125]
There is more wire to travel through,farther distance, and a higher possibility of other disruptions.  Please Mark Brainliest!!!
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3 years ago
Velocity question...is it A B Cor D
Ket [755]
In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

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6 0
3 years ago
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