Answer:
Telescope
Explanation:
Telescope is usually defined as an optical instrument that is commonly used to observe the objects in a magnified way that are located at a large distance from earth. These telescopes are comprised of lenses and curved mirrors that are needed to be arranged in a proper way in order to have a prominent look. It is commonly used by the astronomers.
This was first constructed by Hans Lippershey in the year 1608.
The tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.
<h3>Principle of moments</h3>
The Principle of Moments states that when a body is in equip, the sum of clockwise moment about a point is equal to the sum of anticlockwise moment about the same point.
The formula for calculating moment is given below:
- Moment = Force × perpendicular distance from the pivot
<h3>Calculating the tension in the chains</h3>
From the principle of moments:
Let tension in chain 1 be T1 and tension in chain 2 be T2.
T1 + T2 = 150 + 650 + 419
T1 + T2 =1219
Taking all distances from chain 1,
Sum of Moments = 0
419 × 0.5 + 150 × 0.85 + 650 × 0.9 = T2 × 1.7
T2 = 922/17
T2 = 542.35 N
Then, T1 = 1219 - 542.35
T1 = 676.65 N
Therefore, the tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.
Learn more about tension and moments at: brainly.com/question/187404
brainly.com/question/14303536
If you do this on Earth, then the acceleration of the falling object is 9.8 m/s^2 ... NO MATTER what it's mass is.
If its mass is 10 kg, then the force pulling it down is 98.1 Newtons. Most people call that the object's "weight".
The correct answer would be the sun
Answer:
a. 1222.13 J b. 2.36 m. His new position is above his original position
Explanation:
From work-kinetic energy principle, workdone = change in kinetic energy
So work the skateboarder does on himself W₁ = -116 J
work done by friction W₂ = -264 J
gravitational potential energy change W₃ = mgΔy
The kinetic energy change ΔK = 1/2m(v² - u²) where m = mass of skater = 52.9 kg, u = initial speed of skaterboarder = 2.04 m/s and v = final speed of skaterboarder = 6 m/s. ΔK = 1/2m(v² - u²) = ΔK = 1/2 × 52.9(6² - 2.4²) = 842.13 J
So, W₁ + W₂ + W₃ = ΔK
W₃ = ΔK - W₁ - W₂ = 842.13 J - (-116 J) - (-264J) = 842.13 J + 116 J + 264J = 1222.13 J
b. Since W₃ = mgΔy = 1222.13 J
Δy = W₃/mg = 1222.13/(52.9 × 9.8) = 1222.13/518.42 = 2.36 m
Since Δy > 0, his new position is above his original position.
