Answer:
Explanation:
(a)
Applying the conservation of energy for cube,
i.e the initial kinetic energy of the body = its final potential energy
1/2mv² = mgh
h1 = v²/2g -------------------- (1)
and For cylinder,
the initial kinetic and rotational energy equals the final potential energy
1/2 mv² + 1/2 Iω² = mgh
here I = momentum of inertia of the cylinder
I = 1/2 m R²
∴ 1/2 mv² + 1/2 (1/2mR²)v²/R² = mgh
1/2 mv² + 1/4mv² = mgh
h2 = 3v²/4g -------------- (2)
From equations (1)&(2);
therefore the cylinder will go the greater distance up the incline.
(b)
h1-h2 = v²/2g - 3v²/4g
= 0.25 v²/g
the difference between the maximum distances the objects travel up the incline is 0.25*v²/g.
(c)
here we are taking into account of rotational energy.
One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion.
BTW, circles are ellipses too, but so special that their likelihood is close to zero.
Answer: A Answers. Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes.
Explanation:
What happens to the KE of an object when it slows down and heats up? - Quora. The kinetic energy goes down and the loss of the kinetic energy is through the production of heat energy. In real world this is due to friction, or an opposing force that decelerates the object, or a combination of both.
Answer:
327.58 kPa
Explanation:
Given data; 1st wave
Amplitude ( A1 ) = 293 Kpa
frequency = 6Hz
Average stress = 2388kPa
phase angle = 90° ( leading )
<u>Determine the amplitude of the second sinusoidal component in kPa </u>
Amplitude of additional wave ( A2 ) = 293 / 2 = 146.5 kPa
Amplitude of first wave ( A 1 ) = 293 kPa
hence the amplitude of the second sinusoidal component
A' =
= ![\sqrt293^2 + 146.5^2 + 2(293*146.5) *( -0.45)](https://tex.z-dn.net/?f=%5Csqrt293%5E2%20%2B%20146.5%5E2%20%2B%202%28293%2A146.5%29%20%2A%28%20-0.45%29)
= ![\sqrt{85849+ 21462.25}](https://tex.z-dn.net/?f=%5Csqrt%7B85849%2B%2021462.25%7D)
= 327.58 kPa