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Step2247 [10]
4 years ago
6

List two simple and practical ways in which water can be conserved in the laundry room. In your own words, explain how these str

ategies might have an effect on personal and household water conservation.
Physics
2 answers:
Phantasy [73]4 years ago
5 0
1) Purchase a front-load washer with water/energy conserving features. 
Front load washers use less water then conventional top-load washer, and there are also ones with sensors that even more efficiently manage water usage.

2) Reuse washer water. Many washing machines feature a hose that dumps excess water into a basin during the rinse cycle. This water, rather than going down the drain, could be for your car's first rinse or for other purposes. 
pantera1 [17]4 years ago
4 0
One practical way of conserving water while doing laundry is to separate the used water in a washing machine, put these water in a bucket, and it can be used to flush the toilet. Second, you can used the water to clean your garage and other pavement in front of the house, and can also be use to clean the toilet.
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Which equation represents the total energy of a system?
aev [14]

C) Total Energy = Potential Energy + Kinetic Energy

Explanation:

The total energy of a system (also called mechanical energy) is given by:

E=PE+KE

where

PE is the potential energy

KE is the kinetic energy

The two types of energy have a different origin:

  • Potential energy (PE) is the energy possessed by an object due to its position. It is commonly in the form of gravitational potential energy, which is the energy due to the position of the object in the gravitational field, defined as:

PE=mgh

where m is the mass of the system, g the acceleration of gravity, h the heigth of the object relative to the ground

  • Kinetic energy (KE), which is the energy possessed by an object due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where m is the mass of the system and v is its speed.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0
3 years ago
To drive to the store from home, somebody first travels 5 km west, then they travel 5 km northwest. What is the straight line di
schepotkina [342]
Straight line distance between their home and the store can be solve using cosine law. first is solve the angle which is180 - 45 = 135 degree
C^2 = a^2 + b^2 - 2ab cos(135)c^2 = 5^2 + 5^2 - 2(5)(5) cos(135)c^2 = 85.35c = 9.24 km is the straight line distance between their home and the store
6 0
3 years ago
What happens during the apparent retrograde motion of a planet?
anygoal [31]

Answer:

The planet will move from east to west for a couple of months in the night sky.

Explanation:

Retrograde motion is an optical effect due to the fact that Earth rotates more quickly than the planet that apparently has a retrograde motion in the sky.

For example, Saturn has a slower speed in its orbit around the Sun. That means that the Earth will pass it, and that will give the effect that the planet is moving backward. That same scenario can be seen between two cars on a highway, the faster car will see the slower car when it passes as it is moving for a short fragment of time in backward.

Remember that the planets in the night sky move from west to east, in the case of a planet with retrograde motion, it will move from east to west for a couple of months.

6 0
3 years ago
What is physical quantity​
Brums [2.3K]

A physical quantity is a property of a material or system that can be quantified by measurement. A physical quantity can be expressed as the combination of a numerical value and a unit. For example, the physical quantity mass can be quantified as n kg, where n is the numerical value and kg is the unit.

5 0
3 years ago
Read 2 more answers
The temperature of air changes from 0 to 18°C while its velocity changes from zero to a final velocity, and its elevation change
irina [24]

Answer:

For the air:

Final Velocity 160.77m/s

Final Elevation 1,317.43m

the Internal, Kinetic, and Potential Energy changes  will be equal.

Explanation:

In principle we know the following:

  • <u>Internal Energy:</u> is defined as the energy contained within a system (in terms of thermodynamics). It only accounts for any energy changes due to the internal system (thus any outside forces/changes are not accounted for). In S.I. is defined as U=mC_{V}\Delta T where m is the mass (kg), C_{V} is a specific constant-volume (kJ/kg°C) and \Delta T is the Temperature change in °C.
  • <u>Kinetic Energy:</u> denotes the work done on an object (of given mass m) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as K=\frac{1}{2}mv^2 where v is the velocity of the object in (m/s).
  • <u>Potential Energy:</u> denotes the energy occupied by an object (of given mass m) due to its position with respect to another object. In S.I. is defined as P=mgh, where g is the gravity constant equal to 9,81m/s^2 and h is the elevation (meters).

<em>Note: The Internal energy is unaffected by the Kinetic and Potential Energies.</em>

<u>Given Information:</u>

  • Temperature Change 0°C → 18°C ( thus \Delta T=18°C )
  • Object velocity we shall call it v_{o} and v_{f}, for initial and final, respectively. Here we also know that v_{o}=0m/s^2
  • Object elevation we shall call it h_{o} and h_{f}, for initial and final, respectively. Here we also know that h_{o}= 0m

∴<em> We are trying to find v_{f} and h_{f} of the air where U, K and P are equal.</em>

Lets look at the change in Energy for each.

<u>Step 1: Change in Kinetic Energy=Change in Internal Energy</u>

\Delta E_{K}=\Delta U\\\frac{1}{2}m{v_{f}}^2- \frac{1}{2}m{v_{o}}^2=mC_{V}\Delta T

Here we recall that v_{o}=0m/s^2 and mass m is the same everywhere. Thus we have:

\frac{1}{2}m{v_{f}}^2=mC_{V}\Delta T    

\frac{1}{2} {v_{f}}^2=C_{V}\Delta T\\ {v_{f}}^2=2C_{V}\Delta T\\ v_{f}=\sqrt{2C_{V}\Delta T}     Eqn(1)

<u>Step 2: Change in Potential Energy=Change in Internal Energy</u>

\Delta E_{P}=\Delta U\\mgh_{f}-mgh_{o}=mC_{V}\Delta T

Here we recall that h_{o}=0m/s^2 and mass m is the same everywhere. Thus we have:

mg(h_{f}-h_{o})=mC_{V}\Delta T\\gh_{f}=C_{V}\Delta T\\

h_{f}=\frac{C_{V}\Delta T}{g}      Eqn(2).

Finally by plugging the known values in Eqns (1) and (2) we obtain:

v_{f}=\sqrt{2*718*18}=160.77m/s

h_{f}=\frac{718*18}{9.81}=1,317.43m

Thus we can conclude that for the air final velocity v_{f}=160.77m/s and final elevation h_{f}=1,317.43m the internal, kinetic, and potential energy changes  will be equal.

3 0
3 years ago
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