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shepuryov [24]
3 years ago
13

a horizontal 4-cm-diameter water jet with a velocity of 18 m/s. He impinges the jet normally upon a vertical plate of mass 750 k

g. The plate rides on a nearly frictionless track and is initially stationary. When the jet strikes the plate, the plate begins to move in the direction of the jet. The water always splatters in the plane of the retreating plate. Determine (a)the acceleration of the plate when the jet first strikes it (time
Physics
1 answer:
zmey [24]3 years ago
6 0

Answer:

0.5429 m/s^2

Explanation:

velocity of waterjet = 18 m/s

diameter of water jet ( d ) = 4 cm = 0.04 m

mass of vertical plate(m) = 750 kg

<u>Determine the acceleration of plate when the jet first strikes ( i.e. t = 0 )</u>

first we will determine the impact force

F = β*A*V^2 ----- ( 1 )

where ; β = 1000 kg/m^3  ,  A = π/4 * d^2 , V = 18 m/s

input values into equation 1

F = 407.15 N

<em>finally determine the acceleration at t = 0 </em>

F = m*a

a = F / m =  407.15 / 750 = 0.5429 m/s^2

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

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Answer: Option (B) is the correct answer.

Explanation:

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For example, when a pan is heated and then an egg is placed over the pan. Then there will be transfer of heat from pan towards the egg.

As a result, the egg gets cooked because the egg was touching the pan.

Hence, we can conclude that it is conduction because the egg is touching the pan.

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A sphere of metal has a mass of 1431 g and a diameter of 5.80 cm. What is the density of metal in g/cm3
Dmitriy789 [7]

Answer:

see below

Explanation:

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MariettaO [177]

Answer:

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Energy (E) =?

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E = ½ × 15

E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

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