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Anna [14]
1 year ago
15

Write difference between upstroke and downstroke of water pump

Physics
1 answer:
lakkis [162]1 year ago
7 0

Explanation:

Upstroke is a mechanism which helps to raise the plunger and downstroke helps to help lower the plunger. On the up-stroke of the plunger, the lower valve opens and the upper valve is closed. ... Whereas, on the downstroke, the lower valve closes and the upper one opens.

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Katie rolls a toy car off the end of a table.
hoa [83]

Answer: C.

Explanation:

i choose B and got it wrong it told me the answer after on edge.

4 0
1 year ago
Read 2 more answers
The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a
Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = \frac{excess initial cost}{\Delta M}

payback period = \frac{32 + 66}{520.34}

payback period = 0.188 year

8 0
2 years ago
A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it i
il63 [147K]

Answer:

a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to

This can be calculated by Gauss' Law.

A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.

b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>

8 0
2 years ago
In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat
Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

\frac{Q}{t} = \frac{kA\Delta T}{d}

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

\Delta T = The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

Q_2 = 2*Q_1

Replacing,

kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}

\frac{\Delta T}{x}=2*\frac{\Delta T}{x}

\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}

\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}

Solvinf for T_o,

T_o = -13.18

Therefore the temprature at the outside windows furface when the heat lost per second doubles is  -13.18°C

3 0
2 years ago
A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
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