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2 years ago

The prismatic bar has a cross-sectional area

Physics

1 answer:

ludmilkaskok [199]2 years ago

8 0

solution:

$sum fx=0\\
n=\frac{1}{2}\times\frac{w_{0}}{a}(2a-x)^2\\
n=\frac{w_{0}}{20}(2a-x)^2\\
angle is the normal shus\\
\sigma =\frac{N}{A}\\
\frac{\frac{w_{0}}{2a}(2a-x)^2}{A}\\
\sigma(x)=\frac{w_{0}}{2aA}[2a-x]^2$

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The force required to compress a non-standard spring as a function of displacement from equilibrium x is given by the equation f

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2 years ago

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

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2 years ago

In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves

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Answer:

This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.

We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:

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if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:

I = i*(1, 0, 0)

And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.

To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

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2 years ago

A small car and a large heavier bus are traveling at the same speed. Which has more momentum?

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2 years ago

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Answer:

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3 years ago