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harkovskaia [24]
3 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]3 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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The doppler effect is the change in observed frequency due to
Serga [27]

Answer: relative motion between observer and the sound source.

Explanation: The Doppler effect states that when there is a relative motion between an observer and a sound source the frequency of sound perceived by the observer is different in frequency from the original from the source.

The mathematical back up for this claim is given below.

f' = (v+v') /(v-vs) × f

Where f' = observed frequency

v = speed of sound in air

v' = velocity of observer

vs = velocity of source

f = frequency of sound source.

From the formulae, it can be seen that a change in the value of the velocity of observer (v') and source (vs) produces different value of observed frequency (f').

Note, frequency of sound (f) is a constant.

4 0
4 years ago
Question 1 of 28
amm1812

Answer: D. They are the coldest stars.

Explanation:

5 0
2 years ago
A 4.88 x 10-6 C charge moves 265 m/s
Alex73 [517]

Answer:

F=0N

Explanation:

From the question we are told that:

Charge Q=4.88 x 10-6 C

Velocity v= 265m/s

Angle \theta =0 \textdegree

Magnetic field B=0.0579T

Generally the equation for Force is mathematically given by

 F=Q(\=v*\=B)

 F=qvBsin\theta

Therefore

 F=qvBsin0 \textdegree

 F=0N

5 0
3 years ago
Read 2 more answers
A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she
klasskru [66]
Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer: 
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0
3 years ago
A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature
KiRa [710]

Answer:

A) 854.46 kPa

Explanation:

P₁ = initial pressure of the gas = 400 kPa

P₂ = final pressure of the gas = ?

T₁ = initial temperature of the gas = 110 K

T₂ = final temperature of the gas = 235 K

Using the equation

\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}

Inserting the values

\frac{400}{110}=\frac{P_{2}}{235}

P₂ = 854.46 kPa

3 0
3 years ago
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