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Citrus2011 [14]
2 years ago
9

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a = a

₁ + Fm where a₁ = 3.00 meter/second², F = 12.0 kilogram⋅meter/second² and m =7.00 kilogram. First, which of the following is the correct step for obtaining a common denominator for the two fractions in the expression in solving for a?
a. (m/m times a₁/1) + (1/1 times F/m)
b. (1/m times a₁/1) + (1/m times F/m)
c. (m/m times a₁/1) + (F/F times F/m)
d. (m/m times a₁/1) +(m/m times F/m )
Physics
1 answer:
Nikitich [7]2 years ago
8 0

Answer:

The correct option is option (a).

The acceleration of an object is \frac{33}{7} m/s².

Explanation:

Given expression is

a=a_1+\frac Fm

[ divide a whole number by 1 to turn into a fraction. Since a_1 is a whole number, so  a_1  is divided by 1 to turn into a fraction]

\Rightarrow a=\frac{a_1}{1}+\frac Fm

[The l.c.m of the denominators 1 and m is m. Now multiply both numerator and denominator by m of \frac{a_1}1 and multiply both numerator and denominator by 1 of \frac Fm]

\Rightarrow a=(\frac mm\times\frac{a_1}{1})+(\frac 11\times\frac Fm)

The correct option is option (a)

Given that,

F= 12.0 kg.m/s² , m=7.00 kg and a_1 = 3.00 m/s²

\therefore a=(\frac mm\times\frac{a_1}{1})+(\frac 11\times\frac Fm)

     =(\frac{7.00}{7.00}\times \frac{3.00}{1})+(\frac11\times \frac{12.0}{7.00})

    =\frac{21}{7}+\frac{12}{7}

    =\frac{21+12}{7}

    =\frac{33}{7} m/s²

The acceleration of an object is \frac{33}{7} m/s².

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An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the oc
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Answer:

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Explanation:

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Fw=W+Fi+F

where

Fw=   force or weight produced by the water column above the submarine.

to fint Fw we can use the following ecuation

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2 years ago
A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca
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Answer:

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c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

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v = Image distance = -8 cm

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\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm

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