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ASHA 777 [7]
3 years ago
5

What do the key results indicate?

Physics
2 answers:
svlad2 [7]3 years ago
5 0

Answer:

The key results explain how you'll achieve a certain goal. They have to be quantitative, but always measurable in terms of their progress

Explanation:

Artyom0805 [142]3 years ago
4 0

Answer:

A key result is an important output that then becomes an input to the next key result area, or to the next person. For example, in selling, a key result area is prospecting—finding new, qualified, and interested prospects to talk to about your product or service.

hope this helps!

Explanation:

You might be interested in
A sinusoidal wave of wavelength 2.00m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. At t = 0
Shkiper50 [21]
  1. The frequency and angular frequency are 0.500 Hertz and 3.142 rad/s. respectively.
  2. The angular wave number is equal to 3.142 rad/m.
  3. The wave function for this wave is given by: y = Asin(kx - ωt + Φ).
  4. The equation of motion for the left end of the string is given by: y = 0.100sin(3.142x - 3.142t + 0).
  5. The equation of motion for the left end of the string at x = 1.50 m to the right is equal to y = 0.100sin(4.71 - 3.142t + 0).The maximum speed of any point on the string is 0.3142 m/s.

<h3>How to calculate the frequency and angular frequency?</h3>

First of all, we would determine the frequency of this wave by using this formula:

Frequency = wavelength/speed

Frequency = 0.100/2.00

Frequency = 0.500 Hertz.

For the angular frequency, we have:

Angular frequency, ω = 2πf

Angular frequency, ω = 2 × 3.142 × 0.500

Angular frequency, ω = 3.142 rad/s.

<h3>How to determine the angular wave number?</h3>

Angular wave number, k = 2π/∧

Angular wave number, k = (2 × 3.142)/2.00

Angular wave number, k = 3.142 rad/m.

<h3>How to determine the wave function for this wave?</h3>

Mathematically, the wave function for this wave is given by:

y = Asin(kx - ωt + Φ)

For the equation of motion for the left end of the string, we have:

y = 0.100sin(3.142x - 3.142t + 0)

For the equation of motion for the left end of the string at x = 1.50 m to the right, we have:

y = 0.100sin(3.142x - 3.142t + 0)

y = 0.100sin(3.142(1.5) - 3.142t + 0)

y = 0.100sin(4.71 - 3.142t + 0)

<h3>What is the maximum speed of any point on the string?</h3>

Vy = 0.100sin(- 3.142)cos(3.142x - 3.142t)

Vy ≤ 0.3142 m/s (since cosine varies +1 and -1).

Read more on wave function here: brainly.com/question/11181093

#SPJ4

Complete Question:

A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right.  Initially, the left end of the string is at the origin.  Find:

(a) the frequency and angular frequency,

(b) the angular wave number, and

(c) the wave function for this wave.  

Determine the equation of motion in SI units for

(d) the left end of the string, and

(e) the point on the string at x = 1.50 m to the right of the left end.

(f) What is the maximum speed of any point on the string?

5 0
2 years ago
Someone please help my universe revision!!
Juli2301 [7.4K]
All o the galaxies are moving in a same speed
5 0
3 years ago
I'm thinking A?
Luba_88 [7]

Hey girl

the answer is A

your correct the atmosphere provides warmth


hope this helps

XD


5 0
3 years ago
Compare the gravitational force the sun exerts on Earth to the gravitational force Earth exerts on the sun.
podryga [215]

Answer:

as

  • Mass of sun > Mass of earth

Therefore, the sun will exert more gravitational force on earth.

Explanation:

While comparing the gravitational force exerted by two objects, we need to observe which object has a greater mass.

  • The object with the greater mass exerts a more gravitational force on the other object.

We know that mass of the sun is about 1.99 x 10³⁰ kg, and the earth's mass is only 6.0 x 10²⁴.

as

  • Mass of sun > Mass of earth

Therefore, the sun will exert more gravitational force on earth.

7 0
3 years ago
What would be the best way for her to do this?
andrew-mc [135]
Need more info plz :)


5 0
4 years ago
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