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3 years ago

A jet of water squirts out horizontally from a

Physics

1 answer:

Triss [41]3 years ago

6 0

Answer:

9.54 cm

Explanation:

The water is in free fall after it leaves the tank. The time it takes to land is:

Δy = v₀ t + ½ at²

1.23 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.501 s

Therefore, the horizontal velocity is:

v = 0.685 m / 0.501 s

v = 1.37 m/s

Using Bernoulli equation, where point 1 is the surface of the water and point 2 is the exit of the tank:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since P₁ = P₂, and v₁ = 0:

ρgh₁ = ½ ρ v₂² + ρgh₂

gh₁ = ½ v₂² + gh₂

gh = ½ v₂²

h = v₂²/(2g)

Plugging in:

h = (1.37 m/s)² / (2 × 9.8 m/s²)

h = 0.0954 m

h = 9.54 cm

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3 years ago

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$R=\rho_{L / A}$

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$\begin{aligned}&I=A\left(V / P_{A g} L_{A g}+P_{C u} L C_{U}\right)\\&1=8104 \mathrm{A}\end{aligned}$

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E = resistivity * current density

E = Resistivity * current/area

$E=4.9298 \mathrm{V} / \mathrm{m}$

For silver

$E=4.213 \mathrm{V} / \mathrm{m}$

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For copper

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3 years ago

Which of the following best describes the type of bond formed when sodium reacts with chlorine?

SVEN [57.7K]

Answer: ionic bond

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3 years ago

A tow truck pulls a 2,000-kg car with a force of 4,000 N. what is the acceleration of the car?

ad-work [718]

Answer:

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using f=ma

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3 years ago

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically dr

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Answer:

49.79 m/s

Explanation:

Given:

Initial velocity of the roller coaster is, $u=0\ m/s$

Vertical drop or the displacement of the roller coaster is, $y=-126.5\ m$

The displacement is negative as the motion is in downward direction.

Now, as the motion is in vertical direction only, the acceleration of the roller coaster will be due to gravity acting in the downward direction.

So, the acceleration of the roller coaster is, $a=g=-9.8\ m/s^2$

Now, using the following equation of motion:

$v^2=u^2+2ay$

Where, 'v' is the velocity of the roller coaster at the bottom.

Plug in all the given values and solve for 'v'. This gives,

$v^2=0^2+2(-9.8)(-126.5)\\\\v^2=2479.4\\\\v=\sqrt{2479.4}\\\\v=49.79\ m/s$

Therefore, the speed of the roller coaster at the bottom of the drop is 49.79 m/s.

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3 years ago