Hi:):):):):):):):):):):):):):):):):):):):):):):):)

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

Help with physical science please

alex41 [277]

1. Elastic potential energy (D. EEl)

In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:

$E_{EL}=\frac{1}{2}kx^2$

where k is the spring constant and x is the compression of the spring.

2. Gravitational potential energy (C. Eg)

In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:

$E_g = mgh$

where m is the mass of the toy and g is the gravitational acceleration.

3. Gravitational potential and kinetic energy (A. Eg and EK)

In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by

$E_k = \frac{1}{2}mv^2$

Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.

4. Gravitational potential energy (C. Eg)

The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:

$E_g=mgh$

where m is the mass of the jumper.

5. This exercise has the same text of the previous one.

8 0

3 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$

This means that we have to solve the equation

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0$

Re-arranging it,

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0$

So

$(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m$

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

Why can we never see the backside of the moon from earth?

Gemiola [76]

We can see only 180 degrees ahead.. but any view which is behind that is impossible to see because of that....

5 0

3 years ago

What's the difference between beam balance and spring balance?

seropon [69]

A spring balance will give a different mass reading on the Moon from that on Earth. A beam balance however, will give the same reading.Explain why.

Beam Balance:

Beam balance measures mass.

Mass is the amount of matter the object has.

The S.I. unit of mass is kg


Spring Balance:

Spring balance measures weight not mass.

Disadvantage: It requires gravity to measure.

The S.I. unit of weight is newton (N).

That is why when we are in outer space, we become lighter-there is less gravity!

Mass remains constant while weight is dependent on the gravitational pull of the planet. Mass only changes with a change in matter that results in a change of volume.

Mass remains constant throughout. Only weight changes with gravitational acceleration. Mass will only change with a change in the volume of matter in the body.

5 0

3 years ago