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alukav5142 [94]
2 years ago
11

True or False: Graphs provide a good visual representation of the relationships between the factors investigated in an experimen

t?
Physics
1 answer:
Alecsey [184]2 years ago
8 0

Answer:

True

Explanation:

  • Yes, the graph provides a good visual representation of the relationship between the factors being investigated when there are only two variables being studied at a time.
  • Sometimes a third variable is represented on a negative y-axis which plays a minor role in any one of the factors that may be conveniently represented.
  • Since we can draw only 2-dimensional diagrams on a paper, therefore, graphs are restricted to a couple of variables.
  • The independent variable is usually represented on the x-axis and the dependent variable is represented on the y-axis.
  • Graphs tell us the pattern of variation of a quantity with respect to another.
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Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b
shepuryov [24]

Answer:

  • a. \Delta s ^2 = 8.0888 \ 10^{17} m^2
  • b. \Delta s ^2 = 3.0234 \ 10^{16} m^2
  • c. \Delta s ^2 = 3.0234 \ 10^{20} m^2

Explanation:

The spacetime interval \Delta s^2 is given by

\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2.

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2

\Delta \vec{x}^2 = 5,625 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2

\Delta (c t) ^ 2 = (899,377,374 \ m)^2

\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2

so

\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2

\Delta s ^2 = 8.0888 \ 10^{17} m^2

<h3>b.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2

\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2

so

\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{16} m^2

<h3>c.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2

\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2

so

\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{20} m^2

5 0
3 years ago
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