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Leto [7]
3 years ago
9

Calculate the speed of an 8.0 x 10^4 kg airliner with a kinetic energy of 1.1 x 10^9 J.

Physics
1 answer:
Lynna [10]3 years ago
8 0
KE = ½mv²
1.1 x 10^9 = ½ x 8 x 10^4 v²
v² = 2.75 x 10^4
v ≈ 165.8 m/s

Hope This Helps!
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I’ve been stuck please help !!
Nat2105 [25]

Answer:

The slope of the position time graph gives the velocity.

Explanation:

The slope of the position time graph gives the value of velocity.

In first graph,

The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts.  and more than the second part, so the initial velocity is more than the final velocity.

In the second graph,

The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.

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3 years ago
Under what circumstances will light display total internal reflection?
kumpel [21]
When light is passing from a denser medium to a lighter medium ( for example: from water to air) 
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3 0
4 years ago
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You wish to lift a 12,000 lb stone by a vertical distance of 15 ft. Unfortunately, you can only generate a maximum pushing force
DochEvi [55]

Answer:

5,760,000 lb-ft²/s²

Explanation:

Work done, W = mgy where m = mass of stone = 12,000 lb, g = 32 ft/s² and y = vertical distance = 15 ft.

So, W = mgy = 12,000 lb × 32 ft/s² × 15 ft. = 5,760,000 lb-ft²/s²

6 0
3 years ago
A car turns into a driveway that slopes upward at a 9 degree angle, car is moving at 6.5 m/s. If the driver lets the car coast,
expeople1 [14]

To put it in the simplest form, the automatic transmission has a torque converter that uses the transmission fluid to turn the other side of the torque converter. Ex. Putting two room fans facing each other and turning one on and wind flowing from one fan makes the second fan facing the rotating fan turn and the idling engine probably doesn't have enough power to hold it.

I guess more technically, in an automatic, the car selects a neutral gear (no gear) when it is at rest, and the brake pedal is pressed. Upon releasing the brake, the car will apply a small forward force that will hold the car steady on a very small gradient or propel it forwards slowly on a flat or downhill gradient. This force is only started when the brake is being released, as this is the indicator for the car to change into 1st gear.

In a situation where a car is rolling back down the hill it is facing up, there could be one or more of several situations at hand.

Most likely, it simply lacks power to hold on the gradient of the slope. In this case, you'll have to perform a hill start to maintain brake force until forward propulsion is enough to move the car forward. Hill starts are almost always necessary for manual transmission cars. Additionally, it could be the driver's pedal transfer from brake to accelerator is to slow. That would also account for some part of the slip.

Okay, what's missing here is the clutch. The clutch as I'm sure you're aware is a series of plates that connect to each other and transfer power due to friction. The amount of friction is adjustable depending on how much force is applied to hold the plates together (or apart). It is easier to see this in a manual car at low speeds, where the driver hovers the clutch actuation pedal around the "Friction Point". This point is the fine line between moving (increased friction between clutch plates holds them together more firmly, thus transferring more power) and staying stationary (clutch plates disengaged from each other). From the Friction Point, any further release of the clutch will cause the car to move forwards because the transmission is engaging with more of the engine's power. Depressing the clutch pedal back in will not have any effect, as it will just keep the clutch plates separate.

In an automatic car, this is all controlled by computer algorithms, determining how much the clutch should be engaged to reach a certain speed. Taking off from the lights on a hill for example will not necessarily register as any different to taking off on flat ground. The effect of this is that the car is assuming that is requires a certain number of revs and a predetermined clutch setting to accelerate smoothly. Due to the increased force the hill provides, the car will move backwards until the power again reaches a level that will overcome its slippage.

One other thing, is that clutches work both ways. Since only friction holds them together, the torque exerted by the wheels back through the drive-train to the clutch can cause the plates to slip when they are not completely engaged. This results in the wheels moving independently of the running engine, as the clutch is separating the forces they would exert on each other.

I think the key thing to note is that despite not having a pedal to operate it, Automatic cars still have a clutch - just one that relies on a computer to function.

If it's not the engine or the clutch which are both behaving as normal, and we're still assuming it's an auto, there could be a problem with the car's computer system, transmission, clutch or gearbox that is causing the slippage. I would assume this is less likely, but it might be worth checking if it happens to your car. your frickin' welcome

6 0
3 years ago
Read 2 more answers
How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)
mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0
3 years ago
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