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1 year ago

In a small village in southern Italy, childbirth is considered?

Physics

2 answers:

weeeeeb [17]1 year ago

8 0

Answer:

a community event

Explanation:

Luba_88 [7]1 year ago

8 0

A community event is the right answer

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Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?

Vladimir [108]

Answer:

Explanation:

a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .

First condition of equilibrium is that net force on the body should be zero.

or F net = 0

A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

So first condition of equilibrium is not satisfied.

Second condition is that , net torque acting on the body must be zero.

In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie towards the line joining the body under rotation with the axis .

F is along r

torque = r x F

= r F sinθ

θ = 0 degree

torque = 0

Hence 2nd condition is fulfilled.

7 0

3 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

2 years ago

An object is 15 cm in front of a diverging lens with a

Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}$

$q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm$

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

$h_i = - h_o \frac{q}{p}$

where $h_i, h_o$ are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that $h_i >0$ , which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0

3 years ago

Which properties can be used to determine whether a substance is a liquid or a plasma? Check all that apply.

lapo4ka [179]

Plasma is the state of mater after gases. Its where the gas forms ions. The state is made of charged particles.
So, the best test for plasma is C and D.
The charge of the substance's particles and
The distance between the substance's particles.

6 0

2 years ago

Read 2 more answers

A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find i

serg [7]

Answer:

$\frac{1}{12}ML^2$