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k0ka [10]
3 years ago
5

The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water

is supplied through the hose at 0.006 m3/s and is ejected through the four nozzles into the atmosphere. Determine the constant angular velocity of the arms. Neglect friction at the vertical axis.
Physics
1 answer:
sashaice [31]3 years ago
7 0

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

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A bird flies north 3 kilometers and then south 4 kilometers, what is the resultant displacement of the bird? Do not forget to in
Ber [7]

Answer: let north be+ and south be-

therefore, -1 km  is the resultant displacement of the bird

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8 0
2 years ago
Monochromatic light of a given wavelength is incident on a metal surface. However, no photoelectrons are emitted. If electrons a
mrs_skeptik [129]

Answer:

Light of a shorter wavelength should be used.

Explanation:

This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .

The experiments that have been carried out show that <u>increasing  or decreasing the intensity of the light will not cause the photoelectrons to be emitted</u>, what will cause the photoelectrons to be emitted is to increase the frequency of the incident light.

And a higher frequency corresponds to a shorter wavelength according to the equation:

f=\frac{c}{\lambda}

(where f is frequency, c the speed of light, and \lambda the wavelength)

So the answer is that the wavelength of the light must be shortened to cause the emission of electrones.

4 0
3 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.
IrinaVladis [17]

Answer:

Approximately 722\; \rm m\cdot s^{-1}.

Explanation:

The average speed of a vehicle is calculated as:

\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}.

In this question, the total distance is 2 \times 560\; \rm km = 1120\; \rm km.

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m.

1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m.

Time required for the first part of this trip:

\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s.

Time required for the second part of this trip:

\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s.

The time required for the entire trip would be approximately 930 + 620 = 1550\; \rm s.

Calculate the average speed of this plane:

\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}.

6 0
2 years ago
A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.
horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

7 0
2 years ago
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