Given Information:
Mass = m = 3 kg
Speed = v = 6 m/s
Radius = r = 2 m
Required Information:
Magnitude of the acceleration = a = ?
Answer:
Magnitude of the acceleration = 18 m/s²
Explanation:
The acceleration of the block traveling along a circular path with some velocity is given by
a = v²/r
a = 6²/2
a = 36/2
a = 18 m/s²
Therefore, the magnitude of the acceleration of the block is most nearly equal to 18 m/s².
Bonus:
The corresponding force acting on the block can be found using
F = ma (a = v²/r)
F = mv²/r
Answer:
fluid flowing past the surface of a body exerts a force on it. Lift is the component of this force that is perpendicular to the oncoming flow direction.[1] It contrasts with the drag force, which is the component of the force parallel to the flow direction. Lift conventionally acts in an upward direction in order to counter the force of gravity, but it can act in any direction at right angles to the flow.
If the surrounding fluid is air, the force is called an aerodynamic force. In water or any other liquid, it is called a hydrodynamic force.
Dynamic lift is distinguished from other kinds of lift in fluids. Aerostatic lift or buoyancy, in which an internal fluid is lighter than the surrounding fluid, does not require movement and is used by balloons, blimps, dirigibles, boats, and submarines. Planing lift, in which only the lower portion of the body is immersed in a liquid flow, is used by motorboats, surfboards, and water-skis.
The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = (
- 
) / ( + )
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
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Answer:
If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.
