How would water be different if it wasnt a polar molecule?

Derive an expression for the block's centripetal acceleration ac in terms of m , θ , and physical constants, as appropriate

maxonik [38]

The expression for the block's centripetal acceleration is derived as ω²r or v²/r.

<h3>What is centripetal acceleration?</h3>

The centripetal acceleration of an object is the inward or radial acceleration of an object moving in a circular path.

The expression for the block's centripetal acceleration is derived as follows;

ω = dθ/dt

where;

ω is the angular speed
θ is the angular displacement
t is the time of motion

ac = ω²r

where;

r is the radius of the circular path

Also, ω = v/r

ac = (v/r)²r

ac = v²/r

Thus, the expression for the block's centripetal acceleration is derived as ω²r or v²/r.

Learn more about centripetal acceleration here: brainly.com/question/79801

7 0

2 years ago

Doubling the distance between you and a source of radiation decreases your exposure by:

Elan Coil [88]

Given what we know, we can confirm that doubling the distance between you and a source of radiation decreases your exposure by 75%.

<h3>How is distance related to radiation exposure?</h3>

As expected, increasing the distance from the source of the radiation will reduce its negative effects.
Counter-intuitively however, doubling the distance does not reduce by half, but rather reduces its effects by 3/4th.
This is due to the fact that the radiation effects from the source are inversely proportional to the square of the distance.
This causes the changes to be far greater than expected.

Therefore, given that the radiation is proportional to the square of the distance, instead of being of a more direct relation, we can confirm that when doubling the distance between yourself and the source of the radiation, you can reduce its effects by 3/4 or 75%.

To learn more about radiation visit:

brainly.com/question/9815840?referrer=searchResults

5 0

3 years ago

You are standing on a footbridge that is 12 feet above a lake. You look down and see a duck in the water. The duck is 7 feet awa

kolbaska11 [484]

Answer:

30.22°.

Explanation:

Given that

height of the bridge ,h= 12 ft

The distance of the lake from bridge ,L= 7 m

Lets take angle = θ

Now by using diagram

We can say that

$tan\theta =\dfrac{L}{h}$

$tan\theta =\dfrac{7}{12}$

$tan\theta=0.583$

$\theta=30.22^{\circ}$

That is why the angle will be 30.22°.

6 0

3 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time <em>t</em> :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time <em>t</em> :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$

8 0

3 years ago

If the particle is located slightly to the left of point B, its acceleration is __________.

Igoryamba

(1) directed to the right

Explanation:

To the right of B, u(x) is a decreasing function & so its derivative is negative,this implies that the x component of the force on a particle at this position is positive,or that the force is directed towards right .Small deviations from equilibrium at point B causes a force to accelerate the particle away ,hence particle is in <u>unstable equilibrium.</u>

7 0

4 years ago