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Lady bird [3.3K]
3 years ago
7

Which four equations can be used to solve for acceleration

Physics
1 answer:
emmainna [20.7K]3 years ago
8 0

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as a = \frac{v-u}{t}

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as a = 2*\frac{s-ut}{t^{2} }

So this is second equation for acceleration.

Then from the third equation of motion, v^{2}- u^{2} = 2* a *s

the acceleration equation is determined as a = \frac{v^{2}-u^{2}  }{2s}

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

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A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h
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Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

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the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

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