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Brrunno [24]
4 years ago
6

water jet leaves a horizontal nozzle with a velocity of 12 m/s and mass flow rate of 30 kg/sstrikes the vertical wall of a cart

moving at a fixed direction. After the strike, the water splatters ofin all directions on the surface of the cart wall. Determine the force which must be applied to thecart to: (a) keep it moving at a constant velocity of 4 m/s, (b) to keep it motionless at a fixed posi-tion. Assume that the density of water is 1000 kg/m3.
Physics
1 answer:
amid [387]4 years ago
4 0

Answer:

a) 160 N

b) 360 N

Explanation:

v = 12 m/s

\dot{m} = 30 kg/s

ρ = 1000 kg/m^3

a) when the water  jet is moving at a constant velocity, then force exerted by the jet in the moving plane,

Fx = ρA(v- 4)^2

\dot{m} =ρAv

30 = 1000*A*12

A=2.5*10^(-3)

Fx = ρA(v-4)

Fx = 1000*2.5*10^(-3)*(12 - 4)^2

Fx = 160 N

b) When water jet striking on a stationary vertical plate, the force in the direction of jet:

Fx = ρAv^2

Fx = 1000*2.5*10^(-3)*12^2

Fx = 360 N

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<h2>Answer:</h2>

16°C

<h2>Explanation:</h2>

The equation relating temperature with resistance is given as follows;

R_{T} = R_{0} [\alpha(T^{} - T_{0}) + 1]         --------------(i)

<em>Where;</em>

T_{0} = Temperature at some reference point

R_{0} = Resistance at the reference point

T = Temperature at some other point

R_{T} = Resistance at the other point

\alpha = constant called "temperature coefficient of the resistor material"

<em>From the question; </em>

Let's take the reference point temperature (T_{0}) as 4.00°C;

Therefore;

R_{0} = Resistance at the reference point = 217.4Ω

R_{T} = Resistance at the other point = 216.0Ω

\alpha = Temperature coefficient of the resistor material (carbon) = -0.0005/°C

<em>Now substitute these values into equation (i) as follows;</em>

216.0 = 217.4 [(-0.0005)(T^{} - 4) + 1]

216.0 = 217.4 [-0.0005T^{} + 0.002 + 1]

216.0 = 217.4 [-0.0005T^{} + 1.002]

<em>Divide through by 217.4 as follows;</em>

0.994 = 1.002 - 0.0005T

<em>Solve for T;</em>

0.0005T = 1.002 - 0.994

0.0005T = 0.008

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Therefore, the temperature on a spring day at that resistance is 16°C

4 0
3 years ago
A new system of units is invented where the basic unit of length is the "dak." The "dak" is defined as equal to 3 inches, and th
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Answer:

1 cm = 0.131 daks

Explanation:

It is given that,

1 dak = 3 inches

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We need to find how many "daks" are there in a centimeter. We can do the conversions as follows :

1\ dak=1\ dak\times \dfrac{3\ inches}{1\ dak}\times \dfrac{2.54\ cm}{1\ inch}

1\ dak=7.62\ cm

1\ cm=\dfrac{1}{7.62}\ dak

1 cm = 0.131 dak

So, there are 0.131 daks in 1 centimeter. Hence, this is the required solution.

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4 years ago
A 5.4-kg ball is thrown into the air with an initial velocity of 35.2 m/s.a
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The answer is

Ekin = 1/2 * m * v^2
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Answer:

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For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

F_D =F_g

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V = Velocity

In the other hand we have,

F_g = (m_1 +m_2) g

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(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2

Re-arrange for m_2,

m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1

Our values are given by,

C_d = 0.5

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r = 16.35*10^{-2}m

Replacing in the equation we have,

m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3

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Which three factors are used to calculate gravitational potential energy?
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