<span>Two of them are "redshift" and "blueshift", which are used to describe an object'smotion toward or away from us in space. Redshift indicates that an object is moving away from us. "Blueshift" is a term that astronomers use to describe an object that is moving toward another object or toward us.</span>
Given :
A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.
To Find :
The embankment in the height.
Solution :
Since no external force is acting in the system, therefore total energy will be conserved.
Initial kinetic energy of the object = Energy stored in spring
Also, initial potential energy is 0.
Now,
Therefore, the embankment height is 0.64 m.
Answer:
<h2>3000 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 750 × 4 = 3000
We have the final answer as
<h3>3000 N</h3>
Hope this helps you
The rope will remain taut until the particle makes 79⁰ angle.
<h3>Change in kinetic energy of the particle</h3>
The change in kinetic energy of the particle is calculated as follows;
ΔK.E = K.Ei - K.Ef
Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%
ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E
- let the vertical displacement of the particle = h
- horizontal length = side of the prism = a
- hypotenuse side = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>
K.E = P.E
0.7K.E = mgh
0.7(¹/₂mv²) = mg(Lsinθ)
0.7(v²) = 2g(Lsinθ)
from third kinematic equation;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2g(a tanθ)
0.7(2g(a tanθ)) = 2g(Lsinθ)
0.7(a tanθ) = Lsinθ
0.7a/L = sinθ/tanθ
0.7a/L = cosθ
(0.7 x 0.8)/(3) = cosθ
0.1867 = cosθ
θ = cos⁻¹(0.1867)
θ = 79⁰
Thus, the rope will remain taut until the particle makes 79⁰ angle.
Learn more about kinetic energy here: brainly.com/question/25959744
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Omni-directional. It means all directional.
