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MariettaO [177]
3 years ago
8

Drawing is a form of​

Physics
1 answer:
fenix001 [56]3 years ago
3 0

Answer:

visual art in which one uses to mark drawing and express feelings

Explanation:

In my own opinion. Hope it helps.

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a uniform disc and hollow right circular cone have the same formula for their moment of inertia when rotating about the central
Romashka-Z-Leto [24]

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

I=\int\limits   dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx  = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

6 0
3 years ago
The volume of water in a measuring cylinder is 50 ml. When a piece of stone is immersed
Sergio039 [100]
Volume = 873 - 50 = 723 cm^3
4 0
3 years ago
what is the resistance of a light bulb if a potential difference of 120 v will produce a current of 0.5 a in the bulb? 0.0042 0.
svlad2 [7]
R = V/I

R = 120 V/ 0.5 A

R = 240 ohms
3 0
3 years ago
Read 2 more answers
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
please solve this for me ,A garden roller is pulled with a force of 200N acting at an angle of 50 degree with the ground level.f
Bingel [31]

Answer:

The force pulling the roller along the ground is 128.55 N

Explanation:

A force of 200 N acting at an angle of 50° with the ground level

This force is pulled a garden roller

We need to find the force pulling the roller along the ground

The force that pulling the roller along the ground is the horizontal

component of the force acting

→ The force acting is 200 N at direction 50° with ground (horizontal)

→ The horizontal component = F cosФ

→ F = 200 N , Ф = 50

→ The horizontal component = 200 cos(50) = 128.55 N

128.55 N is the horizontal component of the force that pulling the

roller along the ground

<em>The force pulling the roller along the ground is 128.55 N</em>

8 0
3 years ago
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