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Over [174]
2 years ago
10

A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotat

ing hollow cylinder reaches a certain constant speed, v, the floor moves downward. Both passengers stay "pinned" against the wall of the rotor, as shown in the diagram below.
The magnitude of the frictional force between the adult and the wall of the spinning rotor is F. What is the magnitude of the frictional force between the child and the wall of the spinning rotor?
Physics
1 answer:
Ulleksa [173]2 years ago
8 0

Answer:

 fr ’= ½ F

Explanation:

For this exercise we use the translational equilibrium equation, on the axis parallel to the wall

              fr - W = 0

              fr = W

for the adult man they indicate that the friction force is equal to F

              F = M g

we write the equilibrium equation for the child

             fr ’= w’

             fr ’= m g

in the statement they tell us that the mass of the adult is 2 times the mass of the child

             M = 2m

we substitute

            fr ’= M / 2 g

            fr ’= ½ Mg

we substitute

            fr ’= ½ F

therefore the force of friction in the child is half of the friction in the adult

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A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
Longitudinal sound waves cannot propagate through
strojnjashka [21]

Answer:

A vacuum

Explanation:

Sound waves are examples of mechanical waves. Mechanical waves are waves which are transmitted through the vibrations of the particles in a medium.

For example, sound waves in air consist of oscillations of the air particles, which vibrate back and forth (longitudinal wave) along the direction of propagation of the wave itself.

Given this definition of mechanical wave, we see that such a wave cannot propagate if there is no medium, because there are no particles that would oscillate. Therefore, among the choices given, the following one:

a vacuum

represent the only situation in which a sound wave cannot propagate through: in fact, there are no particles in a vacuum, so the oscillations cannot occur. In all other cases, instead, sound waves can propagate.

3 0
3 years ago
Which statement about the minerals plagioclase feldspar, gypsum, biotite mica, and talc can best be inferred from the chart? (1)
madreJ [45]

Answer:

(4) The physical and chemical properties of these minerals determine how humans use them.

Explanation:

All the materials and the metals found on earth shows certain characteristics naturally or in their physical state. These physical characteristics can be their look, their structure, their color, strength, melting point, boiling point, density, etc.

And chemical properties of the metals are defined as those characteristics or features of the metals that it exhibits when these metals reacts chemically.

The physical properties as well as the chemical properties distinguishes each metal from each other. These properties determines how people use these metals in their life.

7 0
2 years ago
You are riding in an elevator on the way to the
timofeeve [1]

Answer:

F = 104.832 N

Explanation:

given,

upward acceleration of the lift = 1.90 m/s²

mass of box containing new computer = 28 kg.

coefficient of friction = 0.32

magnitude of force = ?

box is moving at constant speed hence acceleration will be zero.

Now force acting due to lift moving upward =

               F = μ m ( g + a )

               F = 0.32 × 28 × ( 9.8 + 1.9 )

              F = 104.832 N

hence, the force applied should be equal to 104.832 N

4 0
3 years ago
A body is accelerated continuously. What is the form of the graph?
Inga [223]
That depends on what quantity is graphed.

It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.

-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces. 
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.

-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.

-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.

-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line. 
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks. 
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.

-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves.  Each piece curves up if speed is increasing, or down if
speed is decreasing.

-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.

6 0
3 years ago
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