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Klio2033 [76]
3 years ago
10

You weigh 730 N.

Physics
1 answer:
JulijaS [17]3 years ago
7 0
The answer is 365N

W = (730N * 2) / 2^2
W = 365N
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Compare and contrast the molecular structure of cleaning bleach and carbon monoxide
dangina [55]
For a very long time, the only real laundry bleach on the market was chlorine bleach, popularized by industry leaders, such as Clorox. Bleach is not only used for stain removal in laundry, but to clean and sterilize objects and surfaces. Chlorine bleach is not good for every fabric and has a very harsh smell, so oxygen bleaches were developed that clean as well as chlorine bleaches in most applications, but are safer on fabrics and are less harsh. Both are effective, but one may be preferable over the other depending on the application.Chlorine Bleach

Chlorine beach is sodium hypochlorite, diluted with water to around a five percent concentration. Manufacturers make it by heating lye (sodium hydroxide) or quicklime (calcium hydroxide) and allowing chlorine gas to bubble up through it. They then add water to the right concentration. Chlorine bleach is highly caustic. It will eat away fabric and skin if left on for an extended period, especially at full strength and take away color. Chlorine bleach is typically diluted even further when used for stain removal or cleaning. It is an unstable product that begins to lose its effectiveness after manufacturing and becomes ineffective over time, and must be stored in a cool, dark place in a plastic container.

Oxygen Bleach

Oxygen bleach is hydrogen peroxide with some sodium and sometimes carbon added to it to form a compound that releases the hydrogen peroxide when added to water. Oxygen leach is a more highly concentrated product than chlorine bleach. Many times, it is found in powdered form, which is then added to water to activate it. Oxygen bleach is known as “color-safe” or “all fabric” bleach, since it does not degrade most fabric or strip most color if used correctly, though you must still test colorfastness before using. It is very stable and can be kept for over a year with no loss of effectiveness. However, it should never be stored in metal or organic containers.

Similarities

Both bleaches work by oxidizing stains and microbes, allowing them to be broken up and lifted away from fabrics and surfaces. Both have excellent anti-microbial qualities that make them good for disinfecting laundry and surfaces, though chlorine bleach has an edge in effectiveness. Neither is effective in cold water, and both require garments be rinsed well after use.

Benefits

Chlorine bleach does not differentiate between color molecules and stains or microbes; it lifts colors away using oxidation as well. Even in low concentrations, it eats away at fabric, so over time, the regular use of bleach will deteriorate garments and fade their color. Chlorine bleach is toxic to aquatic life if released straight into surface water, as in stormdrain runoff from outdoor cleaning projects. It is also harmful to the essential bacteria in septic tanks if used in anything but very small quantities. It works best in hot water, but is also effective in warm water. It cannot be used with other cleaners such as ammonia, as contact can released deadly chlorine gas. It is less expensive to use than oxygen bleach.

Considerations

Oxygen bleach is safe to use on nearly any fabric and to add to laundry loads for extended periods with no damage to clothing. Oxygen bleach turns to water and oxygen when broken down, so it has no negative impact to the environment and is safe for septic systems. It is best if used in the same step as laundry detergent, which makes it even more effective, but combining steps also saves time. It only works well in hot water, but additives can make it effective in warm water.

4 0
3 years ago
I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?
irakobra [83]

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

Original Temperature = 280K

New Temperature = 700K

To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

VT = K

\frac{V1}{T1} = \frac{V2}{T2}

\frac{V1}{T1} = \frac{V2}{T2}

Making V2 as the subject formula, we have;

V_{2}= \frac{V1}{T1} * T_{2}

V_{2}= \frac{2}{700} * 280

V_{2}= 0.0029 * 280

V2 = 0.8L

Therefore, the volume of the gas after it is heated is 0.8L.

7 0
3 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
4 years ago
Two groups of students were tested to compare their speed working math problems. Each group was given the same problems. One gro
oee [108]

Answer:

The people with caculators will probably answer faster due to thier ablitiy to use a device of technology

3 0
4 years ago
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