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dezoksy [38]
3 years ago
8

How is velocity ratio of wheel and axle calculatedaad​

Physics
1 answer:
iogann1982 [59]3 years ago
3 0

Answer:

VR = \frac{Radius of wheel}{Radius of axle}

Explanation:

A machine is a device that can be used to overcome a load by the application of an effort through a pivot. Examples are: pulleys, wedge, screw jack, wheel and axle etc.

The wheel and axle is a simple device that can be used to lift a load through a height. Its velocity ratio (VR) can be determined by:

VR = \frac{Radius of wheel}{Radius of axle}

Note that for a practical wheel and axle, the radius of the wheel is greater than the radius of the axle.

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A 3250-kg aircraft takes 12.5 min to achieve its cruising
scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

n = \frac{P_0}{P}

P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}

kinetic energy= \frac{1}{2} mv^2  =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2

kinetic energy = 90590389.66 kg m^2/s^2

gravitational energy = mgh = 3250*9.8*10000 = 315500000.00  kg m^2/s^2

total energy = 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2

P_o =\frac{409091242.28}{750} = 545454.99 j/s

effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49

effeciency n = = 49%

8 0
3 years ago
A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac
xenn [34]

Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

y-6=-\frac{6}{7}(x-0)

which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

x=-\frac{7}{6}y+7

with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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