Help please it's worth a huge part of my grade!!

Interestingly, there have been several studies using cadavers to determine the moment of inertia of human body parts by letting

loris [4]

Answer:

0.08735 kgm²

Explanation:

m = Mass of lower leg = 5 kg

L = Length of leg = 18 cm

g = Acceleration due to gravity = 9.81 m/s²

f = Frequency = 1.6 Hz

I = Moment of inertia

Time period is given by

$T=2\pi\sqrt{\dfrac{I}{mgL}}$

Also

$T=\dfrac{1}{f}$

So,

$I=\dfrac{mgL}{(2\pi f)^2}\\\Rightarrow I=\dfrac{5\times 9.81\times 0.18}{(2\pi 1.6)^2}\\\Rightarrow I=0.08735\ kgm^2$

The moment of inertia of the lower leg is 0.08735 kgm²

8 0

4 years ago

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass

arlik [135]

Answer:

a) $v_{60^{o}} =4.98 m/s$

b) $\theta_{max}=79.34^{o}$

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

$K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}$

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

$K_{lowest}=K_{60^{o}}+U_{60^{0}}$

So now, we can substitute the respective equations for kinetic and potential energy so we get:

$\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}$

we can divide both sides of the equation into the mass of the pendulum so we get:

$\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}$

and we can multiply both sides of the equation by 2 to get:

$v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}$

so we can solve this for $v_{60^{o}}$ . So we get:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

$cos \theta = \frac{y}{4m}$

so we solve for y to get:

$y = 4cos \theta$

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

$h_{60^{o}}=4m-2m$

$h_{60^{o}}=2m$

So now we can substitute the data in the velocity equation we got before:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

$v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}$

so

$v_{60^{o}}=4.98 m/s$

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

$K_{lowest}+U_{lowest}=K_{max}+U_{max}$

In this case, we know that $U_{lowest}$ will be zero and $K_{max}$ will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

$K_{lowest} =U_{max}$

And now we substitute for the respective kinetic energy and potential energy equations.

$\frac{1}{2}mv_{lowest}^{2}=mgh_{max}$

again, we can divide both sides of the equation into the mass, so we get:

$\frac{1}{2}v_{lowest}^{2}=gh_{max}$

and solve for the height:

$h_{max}=\frac{v_{lowest}^{2}}{2g}$

and substitute:

$h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}$

to get:

$h_{max}=3.26m$

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

$cos \theta_{max}= \frac{0.74}{4}$

$\theta_{max}=cos^{-1}(\frac{0.74}{4})$

so we get:

$\theta_{max}=79.34^{o}$

5 0

3 years ago

A 5.58 kg object with a speed of 35 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate, and rebo

MAVERICK [17]

Answer:

magnitude of vector is 276.19 kg m/s

Explanation:

The initial momentum is vector of magnitude

$5.58 \times 35 = 195.3 (kg m/s)$ And driven in a coherent manner with initial vector.

same magnitude is momentum after the impact, but it is oriented perpendicularly to initial momentum vector.

So, you have 2 momentum vector of specified magnitude perpendicular to one another.

The contrast between such two vectors is a right angle triangle hypotenuse of 195.3 sides

magnitude of vector is $\sqrt{ 195.3^2 + 195.3^2} = 276.19$

8 0

3 years ago

Tom rides his motorcycle at a speed of 15 meters/second for an hour.

Dahasolnce [82]

Answer:

1 hour to ride his motorcycle

4 0

3 years ago

If a tree falls in the woods and nobody is there to hear it, does it really make noise

marshall27 [118]

Ye sit does. But since people weren’t there to hear it, doesn’t mean that it didn’t make a sound.

4 0

4 years ago

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