The diagrams above show an astronaut in a sealed room (with air). The astronaut is weighing a pink vase under three conditions. In diagram A, the room is on the surface of the Earth. In diagram B, the room is floating in space (far from planets, stars, and other gravitational bodies) so that there is effectively zero gravity. In diagram C, the room is also far from gravitational bodies but the room is accelerating upward at 9.8 m/s2.
In diagram B, the scale would read that the weight of the vase is zero. In both diagrams A and C, the scale would read the weight of the vase on the surface of the Earth. In situations A and C, the astronaut would feel exactly the same (with a force equal to the weight of the astronaut between her feet and the floor). The central idea of Einstein's General Theory of Relativity is that there is no experiment the astronaut could do to determine whether she was in situation A or situation C. The two situations are identical. A tremendous amount of evidence corroborating Einstein's idea has been compiled in the last century.
In diagram A, the scale measures the gravitational mass of the vase. On the surface of the Earth, the weight of an object is its gravitational mass multiplied by 9.8 m/s2. In diagram C, the scale is measuring the inertial mass of the vase. The scale would read zero if the vase had no inertia, but since it does have inertia the scale reads the inertial mass of the vase multiplied by its acceleration which is 9.8 m/s2 in diagram C. Another way of stating the central idea of General Relativity is that gravitational and inertial mass are exactly the same. We know of no fundamental reason why this has to be true, but it seems to be true.
The questions that follow explore the effect of inertial mass on the fuzzy dice that your friend has hanging from the rearview mirror of his car. [Note: Air Force pilots in WWII began hanging real dice from the windows of their fighter planes for good luck. That idea evolved into fuzzy dice in automobiles after the war.]
NOTE: Cant really show pictures but they are described
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To answer the questions below, it may be useful to think of your friend's car driving on a level road on the surface of the Earth, or maybe in space accelerating upwards at 9.8 m/s2 (or some other rate of acceleration, depending on the question).
1) Your friend starts out by hanging is fuzzy dice from a spring. On the surface of the Earth, he finds the length of the spring to be 8.4 cm. With his car drifting in space (as in diagram B, above) he finds the length of the spring to be 3.5 cm. What would be the length of the spring in a situation similar to diagram C above, if the car were accelerating upward at a rate of 9.8 m/s2?
2) What would be the length of the spring in a situation similar to diagram C above if the car were accelerating upward at a rate of 12.4 m/s2?
3)What would be the length of the spring in a situation similar to diagram C above if the car were accelerating upward at a rate of 6.6 m/s2?
Note for clarification: Answer all 3 questions since they correspond with each other. and make sure to "bold" the answers somehow. Also just for heads up #1 is NOT 13.3 cm and #2 is NOT 14.6 cm and #3 is not 11.7 cm they were incorrect so don't right those as the answers please. Furthermore, if your answer is wrong then be able to respond back quickly.
To solve these questions, we need to understand the relationship between the length of the spring and the acceleration of the car.
In diagram C, the scale measures the inertial mass of the vase, and the scale reads the inertial mass of the vase multiplied by its acceleration, which is 9.8 m/s^2 in this case.
1) In situation similar to diagram C, if the car were accelerating upward at a rate of 9.8 m/s^2, we can use the same formula to find the length of the spring. Let's call the length of the spring in this situation L1.
Length of the spring in diagram C = Length of the spring on the surface of the Earth × (Inertial mass of the vase × Acceleration)
L1 = 8.4 cm × (Inertial mass of the vase × 9.8 m/s^2)
2) Similarly, for a rate of acceleration of 12.4 m/s^2 in a situation similar to diagram C, let's call the length of the spring L2.
L2 = 8.4 cm × (Inertial mass of the vase × 12.4 m/s^2)
3) For a rate of acceleration of 6.6 m/s^2 in a situation similar to diagram C, let's call the length of the spring L3.
L3 = 8.4 cm × (Inertial mass of the vase × 6.6 m/s^2)
Now, to find the answers, we need to determine the inertial mass of the vase in each case. We know that the weight of an object is its gravitational mass multiplied by the acceleration due to gravity, which is 9.8 m/s^2 on the surface of the Earth.
To calculate the inertial mass, we need to find the ratio of the gravitational mass to the acceleration.
On the surface of the Earth, the weight of the vase is its gravitational mass multiplied by 9.8 m/s^2:
Weight = Gravitational mass × 9.8 m/s^2
By rearranging the formula, we can calculate the inertial mass:
Inertial mass = Weight / 9.8 m/s^2
Inertial mass = Gravitational mass
Now we can substitute the gravitational mass with the inertial mass in the calculations:
1) L1 = 8.4 cm × (Inertial mass of the vase × 9.8 m/s^2)
2) L2 = 8.4 cm × (Inertial mass of the vase × 12.4 m/s^2)
3) L3 = 8.4 cm × (Inertial mass of the vase × 6.6 m/s^2)
Since the equations are the same for all three questions, the only difference lies in the value of the inertial mass (which is equal to the gravitational mass).
To find the value of the inertial mass, we need the given weight or gravitational mass of the vase. Once we have that information, we can substitute it into the equations to obtain the final answers.
Please provide the weight or gravitational mass of the vase, and I will be able to calculate the lengths of the spring in each situation for you.