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Vladimir [108]
3 years ago
9

explain how a roller coaster functions using gravity and momentum. Why does it not need an engine or a motor to speed it up and

slow it down?
Physics
2 answers:
MaRussiya [10]3 years ago
8 0

Answer:

because it will help it go and funcion more

Explanation:

nika2105 [10]3 years ago
3 0

Answer:

It uses graivity and momentum because since we all know, gravity pulls things down, they have a chaint hat lifts the carts up, and then when it gets to a top, it releases it. When it goes down very high, it will go faster. Then it uses momentum when it goes up, then down again, or a loop to regain speed. IT doesn't need a engine or a motor to speed it up or slow it down because gravity and momentum is already doing the job a engine could. Like I said, when it is high in the sky and released, it goes down in incredible speed. And when it needs more speed to keep the ride going, it can add a loop or tight turns to increase it more.

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Which is not a type of driving distraction
Yuliya22 [10]

No grooming your snail while driving. Just ask Spongebob. Oh by the way, what are the options?

7 0
3 years ago
Desde el punto A sale un vehículo a 80 km/h al mismo tiempo sale un ciclista a 20km/h ¿a qué distancia se encuentra uno del otro
stiv31 [10]

Answer:

Distance between them after 5 hours is 300 km.

Explanation:

From point A a vehicle leaves at 80 km / h at the same time a cyclist leaves at 20 km / h at what distance is they from each other after 5 hours.

Distance traveled by A in 5 hours = speed x time = 80 x 5 = 400 km

Distance traveled by B in 5 hours = speed x time = 20 x 5 = 100 km

The distance between them after 5 hours =  400 - 100 = 300 km

4 0
3 years ago
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
8 0
3 years ago
Read 2 more answers
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
Calculate the force generated by a car that hits the wall at an
Makovka662 [10]

This is a defective question. It was WRITTEN by someone who is unclear on the concepts.  DON'T try and answer it.

It's trying to get us to use Newton's second law ... F = m • a.

But that only tells us how much force must act ON THE CAR in order to accelerate it. (45 kg) • (4 m/s^2) = 180 newtons.

This is NOT the force exerted BY the car when it hits something. THAT force depends on its speed WHEN it hits, AND how long it takes for the wreckage to actually come to rest, AND how hard or soft the wall is.

DON'T try to answer this question. Your answer will be wrong, you won't understand why, and the teacher you try to argue with probably won't either.

============================================

More explanation:

Think about jumping off of a ladder in your back yard.  Several times.

Your mass is the same every time.  Your acceleration is the same every time . . . 9.8 m/s² down, the acceleration of Earth gravity, every time.

BUT ...

-- I'll bet you would rather land on wood than on concrete. The force of landing would be less.

-- I'll bet you would rather land on dirt than on wood. The force of landing would be less.

-- I'll bet you would rather land on grass than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a pile of blankets than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a trampoline than on a pile of blankets. The force of landing would be less.

-- I'll bet you would rather jump from a short ladder than from a tall one.  Your speed would be less when you landed, and the force of landing would be less.

==> Your mass is the SAME every time, and your acceleration is the SAME every time.  But the force when you hit is DIFFERENT every time.

The mass and acceleration of the car DON'T tell us the force of the hit when the car hits a wall.  

6 0
3 years ago
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