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liberstina [14]
3 years ago
6

if a truck falls for 33 seconds after it drives off a bridge into canyon below. how far does it fall before it hits the canyon f

loor? give your answer in meters.
Physics
1 answer:
Svetlanka [38]3 years ago
8 0

Here's a formula that's so useful, you should memorize it:

Distance = (1/2) · (acceleration) · (time squared)

On Earth, acceleration of gravity is 9.8 m/s² .

Height of the bridge over the canyon =

Distance the truck falls = (1/2) · (9.8 m/s²) · (33 sec)²

Distance = (4.9 m/s²) · (1,089 sec²)

Distance = 5,336 meters

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I think the correct answer from the choices would be that metals donate electrons to nonmetals. Ionic bonding involves transfer of valence electrons. The metal looses its valence electrons which makes it a cation while the nonmetal accepts these electrons.
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What has been happening to the cosmic microwave background radiation since the Big Bang?​
kondor19780726 [428]

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Explanation:

Cosmologists refer to a "surface of last scattering" when the CMB photons last hit matter; after that, the universe was too big. So when we map the CMB, we are looking back in time to 380,000 years after the Big Bang, just after the universe was opaque to radiation. But the CMB was first found by accident.

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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

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The time it can operate between chargins in minutes is

t=102.8 minutes

Explanation:

Given: m=500kg, r=1.0m, w=200\pi rad/s

a). The rotational kinetic energy

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I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2

K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s

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3 years ago
Hey! I found this question quite interesting. Check it out - https://www.meritnation.com/ask-answer/question/to-raise-a-200kg-st
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6 0
3 years ago
An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign
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Answer:

Option D is correct: 170 µW/m²

Explanation:

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I = c•εo•Eo²/2

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