The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.
<h3>What is Potential and Kinetic energy?</h3>
Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.
mass of rod, mab = 2.4kg
mass of rod, mbc = 4kg
conservation of energy
potential energy at position 1,
V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18
V1 = 11.30112
kinetic energy T1 at position 1 is zero
potential energy at position 2 is zero
K.E at position 2,
= 1/3 *4 * (0.36)²
=0.10368kg m²
= 1/12 *4 * (0.6)²
=0.12kg m²
on putting the values in above equation we get,
T₂ = 1.0667vb²
0 + 11.30112 = 1.0667vb² + 0
vb = 3.2549 m/s
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Answer:
The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.
The second law, $ f(t)=m\,a(t)$ , actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$ . (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$ .)
Newton's third law implies conservation of momentum [138]. It can also be seen as following from the second law: When one object ``pushes'' a second object at some (massless) point of contact using an applied force, there must be an equal and opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.
Explanation:
Answer:
emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.
Direction is counter clockwise.
Explanation:
See attached pictures.
Answer:
Option D, The equator gets more direct sunlight throughout the yea
Explanation:
hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
