Answer:
c. No. An equation may have consistent units but still be numerically invaid.
Explanation:
For an equation to be corrected, it should have consistent units and also be numerically correct.
Most equation are of the form;
(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)
From the above, without the dimensionless constant the equation would be numerically wrong.
For example; Kinetic energy equation.
KE = 0.5(mv^2)
Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.
Take into account that density and relative density are given by:
Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).
The density of water in kg/m^3 is 1000 kg/m^3.
Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:
1 m^3 = 1000000 ml
1 kg = 1000 g
Then, you obtain the following results:
Brass:
Cooper:
In what language are you speaking?
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,
d = V₀t + 0.5gt²
Substituting the known values,
225 = (0 m/s)(t) + (0.5)(9.8)(t²)
Simplifying,
t = 6.776 s
To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.
range = (12.5 m/s)(6.776 s)
range = 84.7 m
<em>Answer: 84.7 m</em>
