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laiz
2 months ago
10

A constant-velocity horizontal water jet from a stationary nozzle impinges normally on a vertical flat plate that rides on a nea

rly frictionless track. As the water jet hits the plate, it begins to move due to the water force. As a result, the acceleration will _____.
Physics
1 answer:
Yuliya22 [3.3K]2 months ago
7 0

Answer:

a = ½ ρ A/M   v₁²

Explanation:

This scenario involves fluid mechanics, where a water jet moving at a constant velocity strikes a paddle and during this interaction, the water remains motionless. We will apply Bernoulli's equation, labeling the jet before the impact as index 1 and after the impact as index 2.

           P₁ + ½ ρ v₁² + ρ g h₁ = P₂ + ½ ρ v₂² + ρ g h₂

In this case, the water is at rest post-impact, hence v₂ = 0, while h₁ = h₂ remains horizontal.

          P₁-P₂ = ½ ρ v₁²

         ΔP = ½ ρ v₁²

Now we can use the definition of pressure as a force divided by area.

         F / A = ½ ρ v₁²

         F = 1/2 ρ A v₁²

The density can be defined as

          ρ = m / V

The volume is

           V = A l

           F = ½ m / l v₁²

With the force known, we can focus on the mass accretion of the plate M leading to its acceleration.

         F = M a

         a = F / M

          a = ½ m/M  1/l   v₁²

           

this can also be expressed in terms of the water's density

          a = ½ ρ A/M   v₁²

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Answer:

2 m/s²

Explanation:

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1 month ago
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A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + \frac{1}{2}at^{2}

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

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We understand that the equation determining the pressure at the base of a fluid column can be expressed through Stevin's law
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