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3 years ago

In a vertical relationship between an employee and supervisor, who must do most of the adjusting?

1 answer:

maks197457 [2]3 years ago

7 0

Wouldn't it be the employee? Because the employee has to adjust to the needs of his/her supervisor. If the supervisor wants 100 boxes the employee has to make those 100 boxes and so on and so forth.

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A left front tire blows out. unless you control the vehicle, what will probably happen?

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You will vere to the left

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3 years ago

the two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different di

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3 years ago

An individual motor disconnect must be in sight of the motor controller and must disconnect the controller. NEC definition of in

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3 years ago

Upper A 12-pound box sits at rest on a horizontal surface, and there is friction between the box and the surface. One side of

Maurinko [17]

Answer:

The coefficient of static friction is : 0.36397

Explanation:

When we have a box on a ramp of angle $\alpha$ , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.

In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:

$F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)$

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

$N=m*g*cos(\alpha)$

and the force of static friction (f) is given as the static coefficient of friction ( $\mu$ ) times the normal N:

$f=\mu *m*g*cos(\alpha)$

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

$f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)$

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

$\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397$

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3 years ago

if the strength of the magnetic feild at a is 200 units and the strength of the magnetic field at b is 50 units what is the dist

IrinaVladis [17]

It would be 100 units.

7 0

3 years ago

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