What about a moving object rarely stays the same?

You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

wlad13 [49]

Answer:

a) V = 0.82m/s

b) Vmax = 0.985 m/s

Explanation:

By conservation of energy we know that:

Eo = Ef $\frac{m*V^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*Xmax$

Solving for V we get:

V = 0.82 m/s

To find the maximum speed we will do the same to an intermediate point where the compression is X and the distance for the work donde by frictions is given by (Xmax - X) = (0.28m - X):

$\frac{m*Vmax^{2}}{2}+\frac{K*X^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*(Xmax-X)$

Then we have to solve for V, derive and equal zero in order to find position X. After solving the derivative we get:

X = 0.1m Replacing this value into the equation for Vmax:

Vmax = 0.985m/s

6 0

3 years ago

Is direction the length of the route between two points ?

zepelin [54]

Answer:

No distance is the length between two routes

Explanation:

Distance is the length of the route between two points. ... Direction is just as important as distance in describing motion. A vector is a quantity that has both size and direction. It can be used to represent the distance and direction of motion.

6 0

2 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms

saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

$\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }$ where $V_{molar} = \frac{A}{\rho }$ an and A is the atomic mass of iron.

Therefore:

$\frac{2}{a^{3} } = \frac{N_{A} * p }{A}$

This implies that:

$A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }$

= $\frac{4}{\sqrt{3} }r$

Assuming that there is no phase change gives:

$\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }$

= 8.60 g/m³

3 0

3 years ago

PLEASE HELP!!! WILL GET BRAINIEST FOR 5 SENTENCES

emmasim [6.3K]

Hello.

BEVs and hydrogen fuel cell vehicles are a more promising transportation technology for the future because they reduce greenhouse gas emissions as well as CO2 making it more 'green.' The second question is your opinion, in mine no, they should not be required but they should at least be considering.
Have a nice day

5 0

3 years ago