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3 years ago

What is the difference between the hegelian right, center, and left? islamic?

1 answer:

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The main difference between<span> the two is that Enlightenment rationalism dwells in abstract inwardness. and it is only through this echo that German Christianity is ...</span>

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A 4.2 m board with a mass of 19 kg is pivoted at its center of gravity. A helium balloon attached 0.24 m from the left end of th

storchak [24]

Answer:

The torque on the board = 11.8 Nm in the anti-clockwise direction

Explanation:

Please find attached the explanation to the answer given.

Download docx

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3 years ago

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

ZanzabumX [31]

The box is moving in the direction of Eduardo with a force of 300N since he is pulling 300N stronger than Clara.

8 0

3 years ago

Read 2 more answers

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

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3 years ago

An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi

MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

$Ft*r=I*a$ -> $(Ft*r)/(a) = I$

Replacing we get that I is:

$I=(200N*0,33m)/(0,936rad/s^2)$

Then $I=70,513kgm^2$

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:

$I=(1/2)*(m*r^2)$

4 0

3 years ago

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

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3 years ago