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
The value of gravitational acceleration at Omicronian in standard units is [tex]2.80 \;\rm meter/second^{2[/tex].
Given data:
The value of gravitational acceleration at planet Omicron Persei 7 is,
[tex]g__{{OP7}}} =7.29 \;\rm \;\rm flurg/grom^{2}[/tex].
We also know that
[tex]1 \;\rm meter = 5.24 \;\rm flurg\\\\\1 \;\rm flurg = \dfrac{1}{5.24} \;\rm meter =0.190\;\rm meter\\\\1 \;\rm grom = 0.493 \;\rm seconds\\[/tex].
Since, the given value of Omicronian gravitational acceleration is in arbitrary set of units. And the standard unit for the acceleration is [tex]m/s^{2}[/tex]. Then converting the arbitrary units to standard units as,
[tex]1 \;\rm flurg/grom^{2} =\dfrac{0.190 \;\rm meter}{0.493 \;\rm second^{2}}\\\\1 \;\rm flurg/grom^{2} =0.385\;\rm meter/second^{2}[/tex]
Now, converting the given value as,
[tex]g__{{OP7}}} =7.29 \;\rm \;\rm flurg/grom^{2} \times \dfrac{0.385 \ meter/second^{2}}{1 \;\rm flurg/grom^{2}} \\\\g__{{OP7}}} =2.80 \;\rm meter/second^{2}[/tex]
Thus, we can conclude that the value of gravitational acceleration at Omicronian in standard units is [tex]2.80 \;\rm meter/second^{2[/tex].
Learn more about the unit conversion from here;
https://brainly.com/question/19420601