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antoniya [11.8K]
4 years ago
11

How to divide 1.08 by 2.7

Mathematics
2 answers:
zaharov [31]4 years ago
6 0

Answer:

0.4

Step-by-step explanation:

Digiron [165]4 years ago
6 0

Answer:

0.5217

Step-by-step explanation:

1.08/2.7

Simplified: 0.5

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A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha
Arlecino [84]

Answer:

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
Two more than 3times and the number is 17 find the number​
Nonamiya [84]

Answer:

5

Step-by-step explanation:

Let the number be x

Therefore 3times the number will be 3x

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5 0
3 years ago
What is the length of segment KR?<br><br> A. 3 mm<br> B. 7 mm<br> C. 8.6 mm<br> D. 17.2 mm
marshall27 [118]

Answer:

Option D. KR=17.2\ mm

Step-by-step explanation:

we know that

The Centroid of a Triangle is the center of the triangle that can be calculated as the point of intersection of all the three medians of a triangle. The centroid divides each median in a ratio of 2:1

In this problem the point R represent the centroid of triangle JKL

therefore

KR=(2/3)KM

substitute the value

KR=(2/3)(25.8)=17.2\ mm

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