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Lapatulllka [165]
4 years ago
9

Finding the spring constant as shown, spring 3, which has an unknown spring constant k3, replaces spring 2. the mass of the weig

ht and pulley are unchanged: m=5.8 kg and mp=1.7 kg . however, because of the different spring constant the distance the pulley hangs below the ceiling is now, h=112.5 cm . the unloaded length of spring 3 is ℓ′3=27.5 cm ; after hanging the mass and pulley, the spring's length is ℓ3=34.5 cm . determine the spring constant of spring 3, k3.
Physics
1 answer:
nevsk [136]4 years ago
6 0
Replaces spring 2. the mass of the weight and pulley are unchanged: m=5.8 kg and mp=1.7 kg
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he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

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

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off
QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

V=\frac{\Delta W}{\Delta q}

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

I=\frac{\Delta q}{\Delta t}

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I is the current (2 A)

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2=\frac{\Delta q}{20}

q=40 C

Then, we can put this value of charge in the voltage equation.

V=\frac{4600}{40}=115 V

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0
3 years ago
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kow [346]
Hey there 

Plate tectonics are found on Earth. 
They are all over the world 

When a plate collides natural disasters happen. 
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5 0
3 years ago
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Answer:

ear

Explanation:

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