Respuesta :
5.72 m/s^2
First, we have available 2 conversion units.
5.24 flurg/m and 0.493 s/grom
I chose the units to describe those two conversions by making the denominator equal to 1 in both cases, so there are 5.24 flurgs per meter and 0.493 seconds per grom.
Now we've been given 7.29 flurg/grom^2 and we want to convert to m/s^2. So we need to figure out how to multiply or divide our conversion factors to cancel out the flurg and grom units. Let's see about cancelling the flurg unit 1st and replacing it with meters
Let's try multiplication
flurg/grom^2 * flurg/m = flurg^2/(m*grom^2)
That won't work. So let's try division
flurg/grom^2 / flurg/m
= flurg/grom^2 * m/flurg
= (m*flurg)/(grom^2 * flurg)
The flurg on top and bottom, cancel, so
= m/grom^2
So dividing by our length conversion will work correctly. Let's do it.
7.29 flurg/grom^2 / 5.24 flurg/m = 1.391221374 m/grom^2
Now we want to convert from m/grom^2 to m/(s grom) using our time conversion factor. Since we want s in the denominator and it's in the numerator, a division looks good. So
m/grom^2 / s/grom
= m/grom^2 * grom/s
= (m*grom)/(grom^2 * s)
= m/(grom * s)
And it is good. So let's do it
1.391221374 m/grom^2 / 0.493 s/grom = 2.821950049 m/(grom s)
We still have one more grom to get rid of. And since it's in the same place as the previous one, let's divide again.
2.821950049 m/(grom s) / 0.493 s/grom = 5.72403661 m/s^2
Since all our input is to only 3 significant figures, round the result to 3 significant figures. Giving 5.72 m/s^2
Unit conversion enables information of a given quantity to be expressed in different units by multiplying the units with the conversion factors
The value of the acceleration due to gravity on the planet Omicron Persei 7, [tex]g_{OP7}[/tex] is approximately 5.724 meter/second²
Reason:
The known parameters:
The acceleration of a freely falling object on the surface of Omicron Persei 7 = [tex]g_{OP7} = \mathbf{7.29} \ \dfrac{flurg}{grom^2}[/tex]
The units used by the architects = [tex]\mathbf{\dfrac{meter}{second^2}}[/tex]
5.24 flurg = 1 meter
1 grom = 0.493 second
Therefore;
[tex]Gravity \ acceleration, \ g_{OP7} =7.29 \ \mathbf{ \dfrac{flurg}{grom^2} \times \left(\dfrac{1 \, meter}{5.24 \, flurg} \right) \times \dfrac{1 \ grom^2}{(0.493 \ second)^2}}[/tex]
[tex]\ \dfrac{flurg}{grom^2} \times \left(\dfrac{1 \, meter}{5.24 \, flurg} \right) \times \dfrac{1 \ grom^2}{(0.493 \ second)^2} = \mathbf{0.7852 \ \dfrac{meter}{second^2}}[/tex]
Which gives;
[tex]Gravity \ acceleration, \ g_{OP7} \approx 7.29 \times 0.7852 \ \dfrac{meter} {second^2} \approx \mathbf{5.724 \ \dfrac{meter} {second^2}}[/tex]
The value of [tex]g_{OP7}[/tex] in SI Units of [tex]\mathbf{\dfrac{meter}{second^2}}[/tex] is therefore;
[tex]g_{OP7} \approx \mathbf{5.724 \dfrac{meter} {second^2}}[/tex]
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