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<span>The relative hardness of a mineral can be tested by the Moh’s scale or the mineral hardness. The Moh’s scale can be characterized by the ability of the mineral to resist scratch resistance by scratching a harder or softer mineral. It has a scale of 1 to 10 where 1 being the softest and 10 being the hardest. The relative hardness of a mineral can best be tested by scratching the mineral across a glass plate. The answer is letter A. </span>

6 0

3 years ago

An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it

steposvetlana [31]

Answer:

μs = 0.36

Explanation:

While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
This acceleration is produced by the centripetal force.
Now, this force is not a different type of force, is the net force acting on the riders in this direction.
Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

$F_{frmax} = \mu_{s}* F_{n} = m * g (1)$

where μs= coefficient of static friction (our unknown)

As we have already said Fn = Fc.
The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

$F_{n} = m* \omega^{2} * r (2)$

Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

$\mu_{s} = \frac{g}{\omega^{2} r} (3)$

Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

$\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)$

Replacing g, ω and r in (3):
$\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)$

3 0

2 years ago

Need help on this thanks​

Need help on this thanks