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Thursday, September 25, 2008
Experience Caned Being
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laboratory experience n + 5
The candle in freefall
Need:
a candle, a match, a transparent and robust.
Design:
a 'further review of the' absence of serious effects in a free fall can be done using a candle, placed in a transparent and robust.
We lit the candle with the match. The flame of a candle is fed by the oxygen of the air: stops in environments (such as municipalities) that is continuously replenished by the convective motions of 'air around the candle flame, which in turn generated by the buoyant force, a consequence of gravity opposed by the container firm. We place the candle in the container and dropped freely container system + candle flame is extinguished, according to forecasts.
Need:
a candle, a match, a transparent and robust.
Design:
a 'further review of the' absence of serious effects in a free fall can be done using a candle, placed in a transparent and robust.
We lit the candle with the match. The flame of a candle is fed by the oxygen of the air: stops in environments (such as municipalities) that is continuously replenished by the convective motions of 'air around the candle flame, which in turn generated by the buoyant force, a consequence of gravity opposed by the container firm. We place the candle in the container and dropped freely container system + candle flame is extinguished, according to forecasts.
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passenger elevator system
Consider a man of 70 kg mass is traveling on an elevator. The motion of the lift is directed vertically and is uniformly accelerated with acceleration a (write all scalar quantities in the game as being directed along the z axis of an orthonormal reference system).
In the scheme of the problem is indicated by the weight of the passenger W (W = 70 kg * 9.80 m / s ^ 2 = 686 N) with R and the force exerted on it from the floor. Note that the third law of motion for R is always equal in magnitude and opposite in direction to the force exerted by the passenger on the floor. The quantities involved are considered positive if it agrees to the z axis drawn in the figure, negative otherwise. Applying the second law of dynamics is written
F = ma, or
R - W = ma, where a is the acceleration
system + passenger lift.
So if for example a = 0.5 m / s ^ 2 (positive, ie upward), we get
R = 70 kg * 9.80 m / s ^ 2 + 70 kg * 0.5 m / s ^ 2 = 721 N,
The passenger, still in the frame of the lift, the feeling of more weight than usual. The 'apparent weight' of the passenger (in the form of the force he exerts on the floor and vice versa) is in fact higher than it would if the elevator was at rest or of uniform motion moved, was that a reference system INS. This could occur by placing the passenger on a dynamometer attached to the floor of the elevator. Where a = -0.5 m / s ^ 2, is
R = 70 kg * 9.80 m / s ^ 2 - 70 kg * 0.5 m / s ^ 2 = 651 N,
The passenger exerts less force on the floor if the direction of acceleration is opposite to the axis z. The example on display is very easy for example reflected in daily life: we can easily realize on the floor to exercise a power greater than our normal weight every time we are in an accelerated upward (lift or games of amusement parks) , the lower when the environment in question is accelerated towards the low.
An interesting case is that of free fall if the lift cable was cut, the system + passenger elevator would fall freely toward the Earth with acceleration a =-g. Then R would be nothing, the passenger and the floor is not exerting any force on each other and the apparent weight of the passenger (possibly indicated by the dynamometer) would be zero.
Consider a man of 70 kg mass is traveling on an elevator. The motion of the lift is directed vertically and is uniformly accelerated with acceleration a (write all scalar quantities in the game as being directed along the z axis of an orthonormal reference system). In the scheme of the problem is indicated by the weight of the passenger W (W = 70 kg * 9.80 m / s ^ 2 = 686 N) with R and the force exerted on it from the floor. Note that the third law of motion for R is always equal in magnitude and opposite in direction to the force exerted by the passenger on the floor. The quantities involved are considered positive if it agrees to the z axis drawn in the figure, negative otherwise. Applying the second law of dynamics is written
F = ma, or
R - W = ma, where a is the acceleration
system + passenger lift.
So if for example a = 0.5 m / s ^ 2 (positive, ie upward), we get
R = 70 kg * 9.80 m / s ^ 2 + 70 kg * 0.5 m / s ^ 2 = 721 N,
The passenger, still in the frame of the lift, the feeling of more weight than usual. The 'apparent weight' of the passenger (in the form of the force he exerts on the floor and vice versa) is in fact higher than it would if the elevator was at rest or of uniform motion moved, was that a reference system INS. This could occur by placing the passenger on a dynamometer attached to the floor of the elevator. Where a = -0.5 m / s ^ 2, is
R = 70 kg * 9.80 m / s ^ 2 - 70 kg * 0.5 m / s ^ 2 = 651 N,
The passenger exerts less force on the floor if the direction of acceleration is opposite to the axis z. The example on display is very easy for example reflected in daily life: we can easily realize on the floor to exercise a power greater than our normal weight every time we are in an accelerated upward (lift or games of amusement parks) , the lower when the environment in question is accelerated towards the low.
An interesting case is that of free fall if the lift cable was cut, the system + passenger elevator would fall freely toward the Earth with acceleration a =-g. Then R would be nothing, the passenger and the floor is not exerting any force on each other and the apparent weight of the passenger (possibly indicated by the dynamometer) would be zero.
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laboratory experience n 4:
free fall of bodies
We verified the absence of apparent weight of a falling body through the law of Stevin. This law states that the difference between the pressures at two points of a fluid in equilibrium in the presence of gravity is given by the pressure exerted at the base by a fluid column height equal to the difference between the two points. We can therefore write: AP = ρgh.
We took a plastic bottle and made a hole a few inches from the bottom. We filled the water bottle so that h is the depth of the hole from the surface of the water and we adjourned to a 'considerable height (eg standing on a table or a chair). The water pressure is p = ρgh, where ρ is its density. We observed a gush of liquid out of the hole, perpendicular to the surface of bottle. We then dropped the bottle freely: the system is the same water bottle + of + passenger elevator system studied in theory. The apparent weight of the water bottle in the frame in free fall is zero, so it's nothing the pressure p calculated according to the law of Stevin. According to forecasts, the gush of water ceases to flow.
We verified the absence of apparent weight of a falling body through the law of Stevin. This law states that the difference between the pressures at two points of a fluid in equilibrium in the presence of gravity is given by the pressure exerted at the base by a fluid column height equal to the difference between the two points. We can therefore write: AP = ρgh.
We took a plastic bottle and made a hole a few inches from the bottom. We filled the water bottle so that h is the depth of the hole from the surface of the water and we adjourned to a 'considerable height (eg standing on a table or a chair). The water pressure is p = ρgh, where ρ is its density. We observed a gush of liquid out of the hole, perpendicular to the surface of bottle. We then dropped the bottle freely: the system is the same water bottle + of + passenger elevator system studied in theory. The apparent weight of the water bottle in the frame in free fall is zero, so it's nothing the pressure p calculated according to the law of Stevin. According to forecasts, the gush of water ceases to flow.
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