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iris [78.8K]
3 years ago
11

Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and

the diameter of the nozzle you hold is 5.19×10−3 m. At what speed v does the water exit the nozzle?
Physics
1 answer:
Tems11 [23]3 years ago
4 0

Answer:

<em>0.153 m/s</em>

<em></em>

Explanation:

The flowrate Q = 0.625 x 10-3 m^3-/s

The diameter of the nozzle d = 5.19 x 10^-3 m

the velocity V = ?

The cross-sectional area of the flow A = \pi d^{2}/4

==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2

From the continuity equation,

Q = AV

V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = <em>0.153 m/s</em>

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Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

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Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other
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A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is
Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration a_{c} is given by the following equation:  

a_{c}=\frac{V^{2}}{r} (1)

Where:  

a_{c}=17 m/s^{2} is the <u>centripetal acceleration</u>

V is the<u> tangential speed</u>

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Isolating V from (1):

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V=\sqrt{(17 m/s^{2})(12 m)}

<u />

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Let two objects of equal mass m collide. Object 1 has initial velocityv, directed to the right, and object 2 isinitially station
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<span>Our equation 1 would be
m*v=M*V1+m*V2
v=V1+V2
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the equation 2 would look like this
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Answer:

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3 blocks toward north = 300 j m

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Displacement is 565.69 m at 45° west of north

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