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harkovskaia [24]
2 years 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:
TEA [102]2 years ago
7 0

Answer:

a = ½ ρ A/M   v₁²

Explanation:

This is a problem of fluid mechanics, where the jet of water at constant speed collides with a paddle, in this collision the water remains at rest, we write the Bernoulli equation, we will use index 1 for the jet before the collision the index c2 for after the crash

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

in this case the water remains at rest after the shock, so v₂ = 0, as well as it goes horizontally h₁ = h₂

          P₁-P₂ = ½ ρ v₁²

         ΔP = ½ ρ v₁²

let's use the definition of pressure as a force on the area

         F / A = ½ ρ v₁²

         F = 1/2 ρ A v₁²

the density is

          ρ = m / V

the volume is

           V = A l

           F = ½ m / l v₁²

knowing the force we can focus on the acceleration of the mass palette M

         F = M a

         a = F / M

          a = ½ m/M  1/l   v₁²

           

as well it can be given depending on the density of the water

          a = ½ ρ A/M   v₁²

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Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0

momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ

Solve the two equations for v₂ and θ.

5 0
2 years ago
Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it
Inessa [10]

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

h_{f}=561.47\ kJ/kg

h_{fg}=2163.8\ kJ/kg

We need to calculate the enthalpy of water at initial state

h_{1}=h_{f}

h_{1}=561.47\ kJ/kg

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

h_{2}=h_{f}+xh_{fg}

Put the value into the formula

h_{2}=561.47+0.5\times2163.8

h_{2}=1643.37\ kJ/kg

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

Q=\dfrac{m(h_{2}-h_{1})}{t}

Put the value into the formula

Q=\dfrac{3\times(1643.37-561.47)}{30\times60}

Q=1.80\ kW

Hence, The average rate of energy transfer to the cooker is 1.80 kW.

3 0
2 years ago
When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large
diamong [38]

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.                  

4 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
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Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

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B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

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A comet is long and also bright, which has a blue-ish color to it. 

Best Regards, Mike
3 0
2 years ago
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