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3 years ago

Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

1 answer:

7 0

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

$Q = VA$

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

$Q_2 = 1.2*10^{-4}m^3/s$

$d = 0.021m$

$r = \frac{0.021}{2} = 0.0105m$

Since there is continuity we have now that,

$V_1A_1 = Q_2$

$V_1A_1 = 1.2*10^{-4}$

$V_1 = \frac{1.2*10^{-4}}{A_2}$

$V_1 = \frac{1.2*10^{-4}}{\pi r^2}$

$V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}$

$V_1 =0.347m/s$

Therefore the speed of the water's house supply line is 0.347m/s

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A light bulb has a voltage of 36 and a current of 8 A. Calculate the resistance of the light bulb

timurjin [86]

R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm

4 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

Why is understanding the concept of forces, friction and gravity important?

nydimaria [60]

Because it's the basis of how everything around you works

8 0

3 years ago

A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1

baherus [9]

Answer:

The value of the power is $P_c = 38.55 \ W$

Explanation:

From the question we are told that

The power rating $P_{1000} =P_b= 52 \ W$

The frequency is $f = 1000 \ Hz$

The frequency at which the sound intensity decreases $f_k = 20 \ Hz$

The decrease in intensity is by $\beta = 1.3 dB$

Generally the initial intensity of the speaker is mathematically represented as

$\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]$

Generally the intensity of the speaker after it has been decreased is

$\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]$

$\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]$

=> $\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3$

=> $\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3$

=> $\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $log_{10} [\frac{P_b}{P_c} ] = 0.13$

taking atilog of both sides

$[\frac{P_b}{P_c} ] = 10^{0.13}$

=> $[\frac{52}{P_c} ] = 10^{0.13}$

=> $P_c = \frac{52}{1.34896}$

=> $P_c = 38.55 \ W$

3 0

3 years ago

In order to analyze a circuit correctly, what values should be known?

Ann [662]

Answer: A

resistances of resistors and voltages of batteries

Explanation:

Electric circuit comprises of resistors or load, batteries or power supplied e.t.c

Current can only flow in a circuit when there are resistors and voltage supply which is a battery.

In order to analyze a circuit correctly, the values of resistances of resistors and voltages of batteries should be known.

3 0

2 years ago