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Mice21 [21]
2 years ago
15

Which expression is equivalent to startfraction startroot 10 endroot over rootindex 4 startroot 8 endroot?

Mathematics
1 answer:
Ludmilka [50]2 years ago
7 0

The expression that is equivalent to √10 / (4√8) is given as √5 / 8

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given the equation:

√10 / (4√8)

Simplifying:

= √10 / (4√(4 * 2))

= √10 / (8√2)

= √5 / 8

The expression that is equivalent to √10 / (4√8) is given as √5 / 8

Find out more on equation at: brainly.com/question/2972832

#SPJ1

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4 years ago
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
a beach ball has a diameter of 2 feet. which of the following is closest to the volume of the beach ball. a.4.2ft b. 12.6ft c. 1
zloy xaker [14]

Answer: A

Step-by-step explanation: v= 4/3 pi r^3

Take diameter to radius by dividing it by 1/2

7 0
3 years ago
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