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Zarrin [17]
3 years ago
15

Form of communication that reaches large audiences without personal contact.

Physics
2 answers:
bagirrra123 [75]3 years ago
8 0

)MASS MEDIUM/MEDIA( which is a form of communication that does not need personal contact to relay or convey and information.

lukranit [14]3 years ago
4 0
Socialization is the process of internalization of the norms and ideologies of society by an individual. It encompasses learning and teaching to attain social and cultural continuity. It also has various forms. One of which is the MASS MEDIUM/MEDIA which is a form of communication that does not need personal contact to relay or convey and information. Examples of which are books, films, internet, magazines and so forth.
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Earth orbits the sun once every
Karo-lina-s [1.5K]

Answer:1 trip around the earth is an angular displacement of 2*pi

3.6525*10^2 days

I

Explanation:24 h/1 day * 3.600*10^3 s/1h = 3.156*10^7 s

Angular speed = angular displacement / time

Angular speed = 2*pi rads / 3.156*10^7 s = 1.9910*10^-7 rad/s

3 0
3 years ago
a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration
slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0
2 years ago
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
2 years ago
A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight
Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is  7.61 x 10⁻⁶  

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

  • m is mass supported by the steel
  • g is acceleration due to gravity
  • A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

  • E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

  • ΔL is change in length
  • L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0
1 year ago
As objects grow farther apart, what happens to the force of gravity between them?
Papessa [141]
It decreses Decreases
8 0
3 years ago
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