What means of motion

A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the

o-na [289]

Answer:

A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.

Gpe @ A=mgh=40*9.81*2=784.8J

B. At 30degree vertical angle the vertical displacement from lowest position is given by

2-2cos(30)=2-1.73=0.27m

Gpe @B= 40*9.81*0.27=106 J

C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.

8 0

3 years ago

When two kinds of pulleys are used together the system is called

jekas [21]

They are called fixed and movable pulleys

3 0

3 years ago

What is the most effective means of establishing awareness of hazards in commercial, industrial, and storage facilities with lar

coldgirl [10]

Answer:

C: Contacting the facilities.

7 0

3 years ago

A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west

Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

$\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}$

(see the attached graphic)

We have

ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or

$\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

wind (relative to Earth) = 5.0 m/s due East, or

$\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)$

ducks (relative to earth) = some speed v due South, or

$\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

Then by setting components equal, we have

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0$

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

We only care about the direction for this question, which we get from the first equation:

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}$

$\cos\theta=-\dfrac57$

$\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)$

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0

3 years ago

Five liters of air at 50 c is warmed to 100c what is the new volume if the pressure remain constant

KonstantinChe [14]

To solve the problem, we can use Charle's law, which states that for an ideal gas at constant pressure the ratio between absolute temperature T and volume V remains constant:
$\frac{T}{V}=k$
For a gas transformation, this law can be rewritten as
$\frac{T_1}{V_1}= \frac{T_2}{V_2}$ (1)
where 1 and 2 label the initial and final conditions of the gas.

Before applying the law, we must convert the temperatures in Kelvin:
$T_1 = 50^{\circ}C + 273 = 323 K$
$T_2 = 100^{\circ}C+273=373 K$
The initial volume of the gas is $V_1 = 5 L$ , so if we re-arrange (1) we find the new volume of the gas:
$V_2 = V_1 \frac{T_2}{T_1}=(5 L) \frac{373 K}{323 K}=5.77 L$

8 0

3 years ago