3 years ago

Select all that apply.

2 answers:

6 0

$\text{Acceleration}=\frac{d^2}{dt^2}\text{displacement}$
So the dimensions of acceleration is $LT^{-2}$
Any answer that comes under that definition is correct.

spin [16.1K]3 years ago

6 0

Strange as it may seem, four of those five choices
are perfectly good units of acceleration.

The only one that isn't is the one in the middle: ft/s

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As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

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