A teacher did an experiment to show the movement of particles in solids, liquids, and gases. The experimental set-up is shown be
low:
The dryer speed can be adjusted at low, medium, and high to represent each state of matter. Which combination of the state of matter and the corresponding dryer speed is correct?
A) liquid = low; solid = medium; gas = high.
B) solid = low; liquid = medium; gas = high.
C) liquid = low; gas = medium; solid = high.
D) gas = low; liquid = medium; solid = high.
The dryer speed can be adjusted at low, medium, and high to represent each state of matter. Which combination of the state of matter and the corresponding dryer speed is correct?
A) liquid = low; solid = medium; gas = high.
B) solid = low; liquid = medium; gas = high.
C) liquid = low; gas = medium; solid = high.
D) gas = low; liquid = medium; solid = high.
1 answer:
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Answer:
The current decreases.
Explanation:
Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is , where V is voltage, I is current and R is resistance.
Rearranging this equation, you get:
and
If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.
Thus, it can be seen that the current will decrease as resistance increases and vice versa.
1) 5.5 N
When the ball is at the bottom of the circle, the equation of the forces is the following:
where
T is the tension in the string, which points upward
mg is the weight of the string, which points downward, with
m = 0.158 kg being the mass of the ball
g = 9.8 m/s^2 being the acceleration due to gravity
is the centripetal force, which points upward, with
v = 5.22 m/s being the speed of the ball
R = 1.1 m being the radius of the circular trajectory
Substituting numbers and re-arranging the formula, we find T:
2) 3.9 N
When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:
And by substituting the numerical values, we find
3) 2.3 N
When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is
And substituting the numerical values and re-arranging it, we find
4) 3.3 m/s
The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:
and re-arranging the equation, we find
There are 3 forces acting on the stoplight:
• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward
• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right
The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that
∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0
We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to
2<em>T</em> sin(<em>θ</em>) = <em>W</em>
2 (1000 N) sin(<em>θ</em>) = 100 N
sin(<em>θ</em>) = 0.05
<em>θ</em> ≈ 2.87°
If <em>y</em> is the vertical distance between the stoplight and the ground, then
tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)
Solve for <em>y</em> :
tan(2.87°) = (15 m - <em>y</em>) / (100 m)
<em>y</em> = 15 m - (100 m) tan(2.87°)
<em>y</em> ≈ 9.99 m
