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Citrus2011 [14]
3 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]3 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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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th
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The angular acceleration when it starting

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The angular acceleration when it stopping:

\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2

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A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
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a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

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U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

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K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

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c)

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E=E_e = 14.3 J

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where h_f is the maximum height reached. Solving for this quantity, we find

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