Flint glass materials
2 answers:
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Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
Work done is by the change in the potential energy of the system. The work done by gravity is 924.63 J.
<h3>What is the Kinetic Energy?</h3>- Potential energy in physics is the energy that an item retains as a result of its position in relation to other objects, internal tensions, electric charge, or other elements.
- The gravitational potential energy of an object, which is based on its mass and distance from another object's center of mass, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field are examples of common types of potential energy. The joule, denoted by the letter J, is the energy unit in the International System of Units (SI).
Solution:
mass = 5.10 kg
height = 18.5 mm
We know that work done by the gravity on the watermelon is the change in the potential energy of the watermelon, therefore,
Work done due to gravity = change in the potential energy of the system
W =
W = mg (h₀ - h₁)
W = 5.10 × 9.8 × 18.5
W = 924.63 J
know more about potential energy brainly.com/question/24284560
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There are 3 forces acting on the stoplight:
• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward
• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right
The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that
∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0
We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to
2<em>T</em> sin(<em>θ</em>) = <em>W</em>
2 (1000 N) sin(<em>θ</em>) = 100 N
sin(<em>θ</em>) = 0.05
<em>θ</em> ≈ 2.87°
If <em>y</em> is the vertical distance between the stoplight and the ground, then
tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)
Solve for <em>y</em> :
tan(2.87°) = (15 m - <em>y</em>) / (100 m)
<em>y</em> = 15 m - (100 m) tan(2.87°)
<em>y</em> ≈ 9.99 m
