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DanielleElmas [232]

2 years ago

15

Constants Periodic Table Suppose the top surface of the vessel in the figure (Figure 1) is subjected to an external gauge pressu

re P2. Derive a formula for the speed, vi, at which the liquid flows from the opening at the bottom into atmospheric pressure, Po. Assume the velocity of the liquid surface, V2, is approximately zero. Express your answer in terms of the variables P2, 41, 42,P, and appropriate constants. v1 = v2/?? + g(12 â y1)] Submit Previous Answers â Correct Figure < 1 of 1 > Part B If P2 = 0.86 atm and Y2 - Y1 = 2.6 m , determine V1 for water. Express your answer using two significant figures. y2-yi mo ÎÎ£Î¦ ?

1 answer:

Gnom [1K]2 years ago

5 0

Answer:

a) v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Water does not flow,

Explanation:

a) For this exercise we will use Bernoulli's equation, where index 1 is at the exit and index 2 on the surface of the water

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0

P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂

½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)

v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)

v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Reduce the pressure to SI units

P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa

P₀ = 1.01 10⁵ Pa

ρ = 1 10³ kg / m³

Let's calculate

v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]

v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]

Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀

For example if the pressure P₂ = P₀

v₁ = √ 5.2

v₁ = 2.28 m / s

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A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei

QveST [7]

gravitational potential energy is given by formula

$U = mgh$

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

$\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}$

$\frac{U_2}{U_1} =\frac{m_2}{m_1}$

given that

$m_2 = 4 m_1$

now we have

$\frac{U_2}{U_1} = 4$

5 0

2 years ago

The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma

Molodets [167]

Answer:

the order of importance must be b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

b e a f c

5 0

2 years ago

A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 2

Luda [366]

Answer:

0.08m/s

Explanation:

Given data

M1= 69kg

v1= 2.61m/s

M2= 0.22kg

v2= 25.2m/s

Before snowball is thrown:

Total mass of skater + snowball = 69+ 0.22 = 69.22kg

Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s

After snowball is thrown:

Let's call the velocity of the skater V.

Total momentum = momentum of skater + momentum of snowball

=69.22V + (5.544)

= 69.22V + 5.544

So:

180.7 = 69.22V+5.544

180.7- 5.544= 69.22V

175.156= 69.22V

V= 175.156/69.22

V = 2.53m/s

The total momentum after catching the snowball is mV or:

(69.0 + 0.22) x V

So:

5.544= 69.22V

V= 5.544/69.22

V=0.08m/s

The velocity of the ice skater after throwing the snowball is 0.08m/s

4 0

1 year ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0

2 years ago

A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .

GREYUIT [131]

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

5 0

2 years ago