What’s the answer??????????????????

I shared a picture of the problem. It’s a basic Physics question and an Algebra question.

julia-pushkina [17]

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

Given the expressions;

$T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}$

Equating both expressions we will have;

$2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}$

Divide both equations by 2π

$\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\$

Square both sides

$(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}$

Take the square root of both sides

$\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}$

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

3 0

3 years ago

Full explanation please I’m asking kindly

-Dominant- [34]

1).

Time = (distance) / (speed)

Time = (1000 meters) / (331 m/s)

Time = 3.02 seconds

2).

Acceleration = (change in speed) / (time for the change)

Change in speed = (second speed) - (first speed)

Change in speed = (50 m/s) - (20 m/s) = 30 m/s

Acceleration = (30 m/s) / (5 seconds)

Acceleration = 6 m/s²

5 0

3 years ago

How does refraction allow thin convex lenses to work?

melomori [17]

Answer:

https://www.khanacademy.org/science/physics/geometric-optics/lenses/v/convex-lenses

Explanation:

Here is a link to a video to tell you about this.

7 0

3 years ago

Read 2 more answers

A ball is thrown straight up in the air. Which of these statements about its kinetic energy is true?A ball is thrown straight up

alexgriva [62]

I believe the correct answer from the choices listed above is option B. The statement that is true about the kinetic energy would be that the ball has the least kinetic energy at the top of its flight. Hope this answers the question. Have a nice day.

7 0

4 years ago

Read 2 more answers

Flapping flight is very energy intensive. A wind tunnel test

steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight= $\frac{P}{m}$

Using the formula

Metabolic power for starting flight= $\frac{12}{0.089}$

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0

4 years ago