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3 years ago

A car slows from 27 m/s to 5 m/s with a constant acceleration for 6.87 s. What is the car’s acceleration?

1 answer:

5 0

In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

a = (Vf-Vo)/ t

a: acceleration (m/s2)

Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s<span>^2</span>

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A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to

kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 = 192 + 56 d + 4d^2 - 192

93 = 56 d + 4d^2

4d^2 + 56 d - 93 = 0

$d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}$

$d=\frac{-56\pm 87.72}{8}$ \

d = 1.5 m

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4 years ago

Consider a uniform hoop of radius r and mass m rolling without slipping. which is larger, its translational kinetic energy or it

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3 years ago

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4 years ago

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Which of the following is required for work to be done on an object?

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I think that in order for work to be done, the object must move in the direction of the force and move over a distance.

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3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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3 years ago