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Jobisdone [24]
3 years ago
5

A 2,000 kg car is parked at the top of a 30 m high hill. What is its potential

Physics
1 answer:
Lunna [17]3 years ago
4 0

So the right answer is of option D<em> </em><em>.</em>

<em>Look </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>

<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>.</em><em>.</em><em>.</em>

<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>

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Read 2 more answers
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
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Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

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The maintenance of fluid balance during exercise. Sodium also promotes retention of ingested fluids and leads to an increased plasma volume response during rehydration. The primary goal of supplementation should be considered, fluid vs carbohydrate provision, and the beverage composition altered accordingly.

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Two forces, F1 = -3000 N and F2 = +1500 N acts on a plane horizontally. The craft is moving with a constant velocity of -850 m/s
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Because the net force must be zero, we conclude that the magnitude of the force is 1500 newtons, and the direction is in the positive axis.

<h3>What is the magnitude and direction of the third force?</h3>

By Newton's laws, we know that if the net force applied to an object is different than zero, then the object is accelerated.

In this case, we know that the object moves with constant velocity, so there is no acceleration, meaning that the net force is equal to zero.

Then we must have:

F1 + F2 + F3 = 0N

Replacing F1 and F2 we get:

-3000 N + 1500N + F3 = 0

F3 = 3000N - 1500 N = 1500N

Then the magnitude of the force is 1500 newtons, and the direction is in the positive axis.

If you want to learn more about Newton's laws, you can read:

brainly.com/question/10454047

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Explanation:

Below is an attachment containing the solution.

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