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cupoosta [38]
3 years ago
10

Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the

pipe with a speed of 2.0 m/s. What is the diameter of the column of water as it hits the bucke g
Physics
1 answer:
cupoosta [38]3 years ago
6 0

The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

\mathbf{v_f ^2 = v_i^2 + 2gh}

\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}

\mathbf{v_f ^2 =151 m/s}

\mathbf{v_f  =\sqrt{151 m/s}}

\mathbf{v_f  =12.29 \ m/s}  

From the equation of continuity:

\mathbf{A_iV_i = A_fV_f}

\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}

\mathbf{ r^2_iV_i =  r^2_fV_f}

\mathbf{ (\dfrac{10}{2})^2\times 2.0 =  r_f^2 \times 12.29}

\mathbf{ 50 = 12.29 \times r_f^2}

\mathbf{ r_f=  \sqrt{\dfrac{50}{12.29} }}

\mathbf{ V_f= 2.02 \ cm }

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

brainly.com/question/10822213

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Kingda Ka has a maximum speed of 57.2 m/s. Determine the height of the hill on this roller coaster. Explain to get full credit,
kodGreya [7K]

This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.

The height of the hill is "166.76 m".

<h3>LAW OF CONSERVATION OF ENERGY:</h3>

According to the law of conservation of energy at the highest point of the roller coaster ride, that is, the hill, the whole (maximum) kinetic energy of the roller coaster is converted into its potential energy:

Maximum\ Kinetic\ Energy\ Lost = Maximum\ Potential\ Energy\ Gain\\\\&#10;\frac{1}{2}mv_{max}^2=mgh\\\\&#10;h=\frac{v_{max}^2}{2g}

where,

  • h = height of the hill = ?
  • v_{max} = maximum velocity = 57.2 m/s
  • g = acceleration due to gravity = 9.81 m/s²

Therefore,

h=\frac{(57.2\ m/s)^2}{2(9.81\ m/s^2)}\\\\&#10;

<u>h = 166.76 m</u>

Learn more about the law of conservation of energy here:

brainly.com/question/101125

7 0
3 years ago
A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?
tester [92]

Answer:

Explanation:

Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.

I = 0.5 mr²

Total kinetic energy, T = 0.5 mv² + 0.5 Iω²

T = 0.5 (mv² + 0.5 mr²ω²)

v = rω

So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²

Rotational kinetic energy is

R = 0.5 Iω² = 0.5 x 0.5 mr²ω²

R = 0.25 mv²

So, R / T = 0.25 / 0.75 = 1/3

5 0
3 years ago
True or false: When an object becomes polarized, it acquires a charge and becomes a charged object.​
PSYCHO15rus [73]

Answer:

i think its true

Explanation:

8 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?
Tresset [83]
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

7 0
4 years ago
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