# How do you solve for a if #F=7 1/3# and #m=1/5# given Newton's second law states the #a=F/m# where a is the acceleration, F is the form, and m is the mass?

To solve for ( a ), we use Newton's second law equation: ( a = \frac{F}{m} ). Plugging in the given values of ( F = 7 \frac{1}{3} ) and ( m = \frac{1}{5} ), we have:

[ a = \frac{7 \frac{1}{3}}{\frac{1}{5}} ]

To simplify, we convert the mixed number ( 7 \frac{1}{3} ) to an improper fraction:

[ 7 \frac{1}{3} = \frac{(7 \times 3) + 1}{3} = \frac{21 + 1}{3} = \frac{22}{3} ]

Now, we can substitute this value into the equation:

[ a = \frac{\frac{22}{3}}{\frac{1}{5}} ]

Next, we multiply the numerator by the reciprocal of the denominator:

[ a = \frac{22}{3} \times 5 = \frac{22 \times 5}{3} = \frac{110}{3} ]

So, ( a = \frac{110}{3} ) or approximately ( a \approx 36.67 ) (rounded to two decimal places).

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force = mass * acceleration.....(a)

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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