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3 years ago

12

Where is potential energy the greatest on a rollercoaster?

1 answer:

lidiya [134]3 years ago

4 0

On a roller coaster, the greatest potential energy is at the highest point of the roller coaster

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A ______ is something that can be placed between the sun and the subject to diffuse the light. *

Gekata [30.6K]

Answer:

scrim

Explanation:

A scrim is something that can be placed between the sun and the subject to diffuse the light.

An instance of a diffuser is a softbox that is put on its front side around a strobe containing diffusion content. The sun is a form of hard light that is often diffused through a scrim. The light rays are dispersed by putting a scrim between the sun and the object, and the harsh sun's rays is gentler.

6 0

3 years ago

Read 2 more answers

Vector sum of 15 m, East and 9 m, West is 24 m, East. True False

Firlakuza [10]

Answer:

False

Explanation:

Because when you go through east

( +x axis ) then you go to west ( -x axis )

You will subtract -9 from +15

it's become +6

( I talk about the displacement not distance) ( West = - East )

I hope that it's a clear ") .

7 0

3 years ago

Read 2 more answers

How many atoms are in 3MgCl2?

Bogdan [553]

Answer:

3MgCl2 has 9 atoms.

Explanation:

The Element Magnesium (Mg) has 3 atoms.

The Element Chloride (Cl) has 6 attoms.

Their fore 6 + 3 is 9 of course. 3MgCl2 has 9 atoms.

BTW: 3MgCl2 is a molecular compound as well as H2O and CO2.

8 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

A home run is hit in such a way that the baseball just clears a wall 21m hgh located 130m from home plate the ball is hit at an

svlad2 [7]

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

3 0

3 years ago