Why is the cathode ray oscilloscope evacuated?

A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi

Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of $12-2x$

Dimension of new box is $(12-2x)\times (12-2x)\times x$

Volume $V=(12-2x)\times (12-2x)\times x$

$V=\left ( 12-2x\right )^2\cdot x$

For maximum volume differentiate with respect to x we get

$\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0$

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

$V=\left ( 12-2\cdot 4\right )^2\cdot 4$

$V=4^3$

$V=64\ in.^3$

6 0

2 years ago

Which of the following statements is correct A. College degree isn't as important as it was in the past b. Over time, I can earn

julia-pushkina [17]

Answer:

I'd say C is the answer they want, though my pedantic side wants to argue for B being true as well.

3 0

3 years ago

A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power

Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

Power is defined as the rate at which work is done;

Power = $\frac{workdone}{time}$

Since the two lifters do the same work at different time, let us estimate their power;

P₁ = $\frac{workdone}{2}$ P₂ = $\frac{workdone }{1}$

We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

The power of the first weight lifter is one-half the second lifter.

4 0

2 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,