Difference between centrifugal and semi-centrifugal clutches

Physics

2 answers:

Bezzdna [24]2 years ago

6 0

Answer:

Semi-centrifugal clutches are used in high powered race cars, to reduce the driver effort. Working is just like semi-centrifugal clutch.

The clutch action is purely under centrifugal force.

At low engine rpm the centrifugal force is low, so there is slip, as engine rpm increases, so does the centrifugal force.

grandymaker [24]2 years ago

4 0

Centrifugal clutch: These are automatic clutches under the action of centrifugal forces. ... Semi – centrifugal clutch: The manual type of clutches which have a pedal assist to it and the clutches becomes free as the system gains rpm time to time. These clutches are used to remove driver effort in racing cars.

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$