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djverab [1.8K]
3 years ago
14

My model is supposed to have

Physics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

yes you can submit

:):):)

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3. Now use the brown knob at the opening of the hose to adjust the angle at which water comes out. When water comes out at an an
Alenkasestr [34]

Answer:

Due to brown knob which controls its angle.

Explanation:

When water comes out at an angle instead of going straight up, water does not shoot as high as before because its speed is controlled by the brown knob in order to throw water in another angle. If we adjust the brown knob present at the opening of hose pipe in a straight angle the water shoot very high because water moves with high pressure and no barrier is present between its way.

4 0
3 years ago
In groupthink, the decision is less important than
Tamiku [17]

Decision is less important than taking in part of discussions.

4 0
3 years ago
How much heat is needed to boil 120 kg of water ?
nekit [7.7K]
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

Hope I helped!! xx


</span>
7 0
3 years ago
2. Which describes the law of reflection?
dmitriy555 [2]

Answer: Choice 1

Law of Reflection

Theta(i)=Theta(r)

Angle of incidence=Angle of reflection

8 0
3 years ago
Read 2 more answers
A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com
bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature =20^{\circ}C

Final Temperature =80^{\circ}C

mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

Initial Geothermal water temperature T_h_i=160^{\circ}C

Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

6 0
3 years ago
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