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forsale [732]
3 years ago
8

Which situation best describes the act of reducing? paying bills online instead of sending paper through the mail donating gentl

y used clothes and household items using plastic bags a second time to avoid buying new ones buying products that can be recycled so they don’t end up in the trash
Physics
2 answers:
malfutka [58]3 years ago
6 0

Answer: The correct answer is-

paying bills online instead of sending paper through the mail.

Act of reducing can be described as a behavior/ an act that tends to reduce or decline the usage/ amount of particular thing.

Out of the given options, paying of bills through online mode instead of sending papers through mail best describes the act of reducing.

For instance, a recruitment agency such as CSIR ( Council of scientific and industrial Research) ask for online applications instead of sending them hard copy.

This will reduce the amount of paper as all the applicants will apply through online mode.

Thus, option first is the right answer.

Ugo [173]3 years ago
3 0
In my opinion, i think the first one is the best one since we're cutting down on paper wasted in mails when its better to do it online.

"paying bills online instead of sending paper through mail"
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Which one of the following temperatures (in °C) is equivalent to 294 K?
vaieri [72.5K]

Answer:

21

Explanation: its actually 20.85 but i guess they round to 21

5 0
2 years ago
You could use an elevator or the stairs to lift a box to tenth floor.which has greater power?
erastova [34]

Stairs don't have any power at all.  All the energy used to climb them
has to come from your muscles.

An elevator gets its power from the electric motors that lift it.  All YOU
have to do is stand there and look around.

All of this is a big part of the reason why elevators have become so
popular, and why no buildings with more than a few floors were built
before elevators were invented.


7 0
2 years ago
Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s
Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19=  -5.625 10^35 eV

the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
8 0
2 years ago
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

3 0
1 year ago
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
2 years ago
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