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sukhopar [10]
2 years ago
12

(YOU WILL GET BRAINLIEST)Matter may be classified as a pure substance or a mixture. Where on the Venn diagram would you insert t

he term "pure substance"?
A) A
B) B
C) C
D) D

Physics
2 answers:
inessss [21]2 years ago
3 0
I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms
Klio2033 [76]2 years ago
3 0

Answer: Option (B) is the correct answer.

Explanation:

Anything that has space and occupies volume is known as matter.

A pure substance can be molecule of an element or molecule of a compound.

A molecule can have atoms of either same or different element. Whereas a compound has atoms of different elements.

For example, H_{2} is a molecule with similar atoms, that is, hydrogen element.

H_{2}O is a compound that contains hydrogen and oxygen atom.

Thus, we can conclude that pure substance contains molecule of an element or molecule of a compound.

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In this 2-dimensional graph, the x-component of each vector is the horizontal distance from the origin, while the y-component of each vector is the vertical distance from the origin. It can be seen that the c vector is 1 vertical unit away from the origin, which means that it has a y-component of 1.
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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

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if forces acting on an object are unbalanced, which factors may result from an unbalanced force? Check all that apply. The net f
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The net force is negative, and there is a change in motion.
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2 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti
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The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

\sum F = ma (1)

where

\sum F is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

a=1.50 m/s^2

And there are two forces acting on the child+cart system:

  • The forward force of pushing, F
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Therefore we can write the net force as

\sum F = F -R

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

F-R=ma

And solving for F,

F=ma+R=(30.0)(1.50)+15.0=60 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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