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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$

Now we can find the linear speed of the ladybug, which is given by

$v=\omega r$

where:

$\omega=6.28 rad/s$ is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

$v=(6.28)(0.65)=4.1 m/s$

Learn more about angular speed:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0

4 years ago

A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$

3 0

4 years ago

How do quantum numbers relate to electrons? please explain?

alukav5142 [94]

Quantum numbers allow us to both simplify and dig deeper into electron configurations. Electron configurations allow us to identify energy level, subshell, and the number of electrons in those locations. If you choose to go a bit further, you can also add in x,y, or z subscripts to describe the exact orbital of those subshells (for example 2px). Simply put, electron configurations are more focused on location of electrons then anything else.

Quantum numbers allow us to dig deeper into the electron configurations by allowing us to focus on electrons' quantum nature. This includes such properties as principle energy (size) (n), magnitude of angular momentum (shape) (l), orientation in space (m), and the spinning nature of the electron. In terms of connecting quantum numbers back to electron configurations, n is related to the energy level, l is related to the subshell, m is related to the orbital, and s is due to Pauli Exclusion Principle.

7 0

3 years ago

Thermodynamics

Marizza181 [45]

Answer:

A heat engine is a device that converts internal energy into work. Internal energy is increased by the addition of heat. The efficiency of a heat engine is a measurement of how efficiently it works. Efficiency compares the amount of useful energy extracted from a process to the total energy input. The heat engine will be more efficient if the percentage is higher.

Explanation:

5 0

3 years ago

An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan

DanielleElmas [232]

Answer:

0.872m/s

Explanation:

Tangential velocity is given by the formula,

$v= 2\pi r/ t$

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

$\frac{1lap* 2 *3.14 *25m}{180secs}$

v = 157m/180secs

Therefore, the magnitude of the tangential velocity

=0.872m/secs

5 0

3 years ago