When is the world going to end? Give me your opinions and the reasons why you think that.

The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i

gregori [183]

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

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:

$\int\limits^a_b$ γhxdy

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:

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

6 0

3 years ago

What will happen to the mass and volume of a block if their is two blocks, three, blocks or four blocks? For example, if the mas

Crazy boy [7]

To solve this problem, it would be helpful to know the density of 1 block Density is defined as the mass of the substance per volume. From the example given, The density of the block is (7g)/(15.625 units^3) or 0.448 g/units^3. So, if a block is added, the new mass is 7g + 7g = 14 g And the volume 14 g /(density) = 1 unit^3

7 0

3 years ago

estimate the force that a cyclist travelling at 6m/s needs to apply to the brakes to bring the bicycle to a stop in a distance o

RSB [31]

v1 = 6m/s

v2 = 0

∆v = v1 - v2 = 6m\s

s = t * v = 15m

t = s\v1 = 15(m) \ 6(m\s) = 2.5s

a = ∆v\t = 6(m\s) \ 2.5s = 2.4m\s2

a = F\m = 2.4m\s2

F = a * m = 2.4m\s2 * ?kg