When a piston is at its lowest position in the cylinder, it's said to be at A. BDC. B. TDC. C. BHP. D. IHP.

A young girl slides down a rope. As she slides

hodyreva [135]

Answer:

The girl will move with constant velocity

Explanation:

If after a certain time t_0 the velocity of the girl is v_0 =gt_0 and the upward force on the girl due to rope is mg ,where g is gravitational acceleration. Then the girl will move down with the constant velocity v_0 .

The girl will move with constant velocity,as explained above.

6 0

3 years ago

In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

$V_{f}=a_{ave} \Delta t$

$V_{f}=(1.80 m/s^{2})(85.6 s)$

Finally:

$V_{f}=154.08 m/s$

8 0

3 years ago

A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

So

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

6 0

3 years ago

Based on Schaffer's arguments in "The Value of a Sherpa Life," choose the statement

Tju [1.3M]

Answer:

Sherpas do work that is much more meaningful than the work other climbers do.

Explanation:

3 0

3 years ago

Unpolarized light with an average intensity of 845 W/m2 enters a polarizer with a vertical transmission axis. The transmitted li

RideAnS [48]

The concept to develop this problem is the Law of Malus. Which describes what happens with the light intensity once it passes through a polarized material.

Mathematically this can be expressed as

$I = I_0 cos^2\theta$

Where

I = New intensity after pass through the Polarizer

$I_0$ = Original intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

When the light passes perpendicularly through the first polarizer, the light intensity is reduced by half which will cause the intensity to be $225W / m ^ 2$ at the output of the new polarizer, mathematically:

$I= \frac{I_0}{2} cos^2\theta$

$225 = \frac{845}{2}cos^2\theta$

Solving to find the angle we have

$\theta = 43.11\°$

The orientation angle of the second polarizer relative to the first one is 43.11°

5 0

3 years ago