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Alinara [238K]
3 years ago
14

Any mechanical device which is used to help people do work is a .a0

Chemistry
2 answers:
Leokris [45]3 years ago
8 0

Any electronic device would be a robot, computer, or machine.

KIM [24]3 years ago
3 0
Any mechanical device which is used to help people is a (machine)
the answer is machine
You might be interested in
Instructions
ivann1987 [24]

Answer:

I got a 100 with this, sorry if this is not what you want just trying to help

Explanation:

1. This experiment was to find how mass and speed effect KE. This is important because if you were in a situation where you needed something to go higher, you would know to add more or less of mass/speed.  

To test mass, we filled the bean bag with a certain amount of water, then dropped it. After, you recorded how high it made the bean bag go. The same with speed, but same amount in the bottle, just dropped from different heights.  

My hypothesis is when you have more mass, the KE will be greater. This is also the same with speed, if it is dropped from a higher place, the bean bag will launch farther than the last time.  

2. Data I collected from the lab was like my hypothesis explained. When the height of the bottle increased, it made the bean bag go higher than the last. And I tested 4 different masses, 0.125 kg, 0.250kg, 0.375kg and 0.500kg. Each time the bean bag went higher on a larger mass.  

A lot of times on the speed test, the bean bag would go higher than the bottle drop point, but not every time. Also, when it was dropped from the same height each time, some results varied quite a bit, like when it was dropped from 1.28 the results were 1.14 then 1.30 1.30. Mass on the other hand was all in the same number range, only once the numbers were a bit off from each other.  

3.  Some formulas I used were KE= ½ mv^2 and Ht v^2/2g. The first was to calculate the kinetic energy of an object, m=mass v=speed. Second was for finding out what height I needed to drop something to reach a certain speed, Ht=Height and g= Gravitational Acceleration of 9.8 m/s^2.  

I used these to figure out tables that showed relationships between different things like mass and KE or speed and height. The whole time I was doing the lab, my data was going up, when there was more mass/speed there were higher values in the table.  

This means that my hypothesis at the beginning was correct, more of m/s means KE will increase proportionally because they are all linear. I found it surprising when the bean bag height went over the water bottle drop mark.  

4.     To conclude, my hypothesis matched my data. The data values went up when more mass or speed was added. This means if I were in a situation where I needed more kinetic energy for something, I would know to increase mass or the speed of the object giving it energy.  

The reason that this hypothesis is correct is when you have more mass, you have more energy. So, when you drop let's say a baseball, it isn’t that heavy so it would only launch the bean bag so far. But a bowling ball is very heavy and has lots of energy when falling because of that, it would make the bean bag go very high.  

To make this experiment better, I would use a smoother material for the lever so energy wouldn’t be lost by friction from wood rubbing together. Also, maybe a scanner or video camera to more accurately record how far the bean bag went. All of these would help the lab get more precise results, maybe they could be used in a future lab.

8 0
3 years ago
Help!!
anyanavicka [17]

I believe this a PV = nRT question whereas

you re write the formula and solve for volume

V = nRT/ P

then you input the values

P= pressure constant

V= x

n = moles = 0.2540

R = gas constant should be 8.314J mol

T = C degrees + 273.15 = K

solve for voume

make sure all units match

and use sig figs!!!!

7 0
3 years ago
At a certain temperature the rate of this reaction is second order in NH4OH with a rate constant of 34.1 M^-1s^-1: Suppose a ves
Tom [10]

Answer:

Time = 0.929s = 0.93s (2 s.f)

Explanation:

Rate constant, k = 34.1 M^-1s^-1

Initial Concentration, [A]o = 0.100M

Time = ?

Final Concentration [A] = 0.0240M

The parameters are represented in the following equation as;

1/[A] = kt + 1/[A]o

kt = 1/[A]  -  1/[A]o

kt = 1/0.0240 - 1/0.1

kt = 31.67

t = 31.67 / 34.1

t = 0.929s = 0.93s (2 s.f)

5 0
3 years ago
Wine goes bad soon after opening because the ethanol ch3ch2oh in it reacts with oxygen gas o2 from the air to form water h2o and
lilavasa [31]
The balanced chemical equation of the reaction described above is,

                            C2H6O + O2 --> H2O  + C2H4O2

If we have 3.84 g of oxygen, we divide by its molar mass.
 
                               n = (3.54 g Oxygen gas) x (1 mole O2/ 32 g O2)
                                 n = 0.11 moles O2

Using ratio and proportion,

               number of moles of ethanol = (0.11 moles O2) x (1 mole C2H6)
                                                  = 0.11 moles C2H6

Then, we multiply the calculated value to its molar mass, 46 grams /mol.
                    mass of ethanol = (0.11 mol) x (46 grams / mol)
                                                = <em>5.06 grams</em>
8 0
3 years ago
What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of
Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 =  -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m=  molality

molality = \frac{moles of solute}{mass of solvent in kilograms}

putting the value in the equation:

molality= \frac{1.31}{0.53}

             = 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

6 0
3 years ago
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