7. MANSON'S FAVORITE SINGING GROUP WAS...

Another word for an amp-volt is a(n) a.ohm b.watt c.joule d.coulomb

creativ13 [48]

Power = (voltage) x (current)

1 watt = (1 volt) x (1ampere)

= 1 "volt-amp" or 1 "amp-volt" .

4 0

3 years ago

An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. The car travels to the bottom, through a loop, an

aalyn [17]

Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.

Potential energy = (M) (G) (H) = (800) (9.8) (45) = 352,800 joules

It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.

Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules

So the coaster starts out with (352,000 + 6,400) = 359,200 joules of energy.

There's no friction, so it'll have that same energy at every point of the story.
=================================

Skip the loop for a moment, because the first question concerns the hill after
the loop. We'll come back to it.

The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is

(1/2) (M) (V)² = (400) (10)² = 40,000 joules.

Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.

PE = (M) (G) (H)

319,200 = (800) (9.8) (H)

H = (319,200) / (800 x 9.8) = 40.71 meters
=================================

Now back to the loop:

You said that the loop is 22m high at the top. The PE up there is

PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules

So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.

KE = (1/2) (M) (V)² = 186,720

(400) (V)² = 186,720

V² = 186,720 / 400 = 466.8

V = √466.8 = 21.61 m/s
===============================

Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.

At 4m off the ground, PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.

The rest will be kinetic. KE = (359,200 - 31,360) = 327,840 joules

KE = (1/2) (M) (V)² = 327,840

400 V² = 327,840

V² = 327,840 / 400 = 819.6

V = √819.6 = 28.63 m/s at the end
=======================================

If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .

If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere. Sorry.

6 0

3 years ago

Air will break down (lose its insulating quality) and sparking will result if the field strength is increased to about 3 × 106 N

CaHeK987 [17]

Answer:

Explanation:

Given

Electric Field Strength $E=3\times 10^6\ N/C$

When a charged particle enters the Electric field it Experience a force in the direction of electric field (for positive charge and vice-versa for negative charge)

charge on Electron $q=1.6\times 10^{-19}\ C$

mass of electron $m=9.1\times 10^{-31}\ kg$

Force $F=q\times E$

$F=1.6\times 10^{-19}\times 3\times 10^6$

$F=4.8\times 10^{-13}\ N$

acceleration of electron $a=\frac{F}{m}$

$a=\frac{4.8\times 10^{-13}}{9.1\times 10^{-31}}$

$a=0.527\times 10^{18}\ m/s^2$

6 0

3 years ago

How can you tell which substances are creating the light of the sun or a lightbulb?

Westkost [7]

Through spectroscopy. Each atom in a gas absorbs and emits light at very specific and unique frequencies. Heating up a gas causes it to glow at these frequencies. If you put the light from a mercury lamp, for example, through a prism there will not be a rainbow. There will only be specific bands of light at certain colors.
On the other hand, white light comes from the sun. The inner part of the sun creates white light as it isn't just a gas state (this specific choosing of frequencies is a gaseous phenomenon) but the atmosphere of the sun is a gas. So when the white light passes through it, it absorbs specific frequencies specific to the elements in the gas. These get scattered (released at random directions) and so many of them don't reach our telescopes. So in a rainbow from stars (including the sun) have dark bands at specific frequencies. You need equipment to focus and see the spectrum closely to notice this. But the missing frequencies are the EXACT frequencies that the gas of the sun's atmosphere would release if heated in a lamp.
So based on what light is emitted (in gas bulbs) or missing from a spectrum (from stars) we can tell what elements are present there.

4 0

3 years ago

7. If I have a first-order pressure meter with a time constant of 5s that is held in a chamber at 100 psia, how long will it tak

Alex Ar [27]

Answer:

WOW ITS SO JUUUUIIICCCYY

Explanation:

U USED TO CALL ME ON MY POOPPHNE LATE NIGHT WHEN I NEEDED URRR POOOOPPP

7 0

2 years ago