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kotykmax [81]
3 years ago
10

I don't know how to answer this please help this is urgent.

Mathematics
1 answer:
lukranit [14]3 years ago
7 0

Answer:

V = 240 in ^3

SA = 2(lw+lh+wh)

SA = 236 in^2

Step-by-step explanation:

Volume = l*w*h where l=length, w= width and h =height

We know l=6, w=5 and h =8

V = 6*5*8

V = 240 in ^3

The formula for surface area of a rectangular prism is

SA = 2(lw+lh+wh)  where l=length, w= width and h =height

We know l=6, w=5 and h =8

Substituting into the equation

SA = 2(6*5+6*8+5*8)

     = 2 (30+48+40)

     =2(118)

     = 236 in ^2

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A) F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }with a = \frac{2\Delta x}{t^2}

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha) (1)

where \alpha is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration a that can be calculated as a function of the given displacement and time interval as:

a = \frac{2\Delta x}{t^2} and the force equation is:

\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma

now substituting N from equation (1):

F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}

and putting the expression for the acceleration:

\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}} (2)

which is the requested expression

B) Substituting the values in (2) and using g=9.8\ \rm{ms^{-2}}:

F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}

The gravitational force acting on the box is:

F_g=mg=50\cdot9.8=490\ \rm{N}

F/F_g = (117.634/490)\cdot 100 = 24 \%

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.

8 0
4 years ago
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