Doe the baby diaper tabs go to the front or back. Huggies diaper.

Physics

2 answers:

alukav5142 [94]4 years ago

7 0

Answer:

Front of the diaper

You take the straps that are connected to the back of the diaper and pull the. forward to attach to the front.

hope this helps.

Zielflug [23.3K]4 years ago

3 0

Answer:

back lol

Explanation:

Pull the tabs and wrap around to stick to the front of the diaper:)

You might be interested in

7. What can be done to prevent such devastating destruction as seen in the earthquake in Haiti? correct me if im wrong on the su

Alex787 [66]

Well, there would have to major supports on every building that was tall even though we have very strong foundation the foundation doesn't do anything except no give us dirt as a floor.but a really strong structure to use is a triangle formation.
Hope this helped

3 0

3 years ago

True or false? Once the sun explodes, the whole solar system will go with it.

liberstina [14]

True. That is how powerful the sun is. But it will explode in only about 5 billion years.

6 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$