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zhenek [66]
3 years ago
14

If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?

Record to the ones place.
______cm3
Determine the initial volume of water in the graduated cylinder. If you added the black marble to the graduated cylinder and it sinks, what final volume should the water level indicate? Record to the ones place.
______ML

Physics
2 answers:
Mkey [24]3 years ago
7 0

1. Answer: 14.137{cm}^{3}


Assuming the marble has an spherical shape, its volume can be calculated by the following formula:


V_{sphere}=\frac{4}{3}\pi{r}^{3}     (1)


where r is the radius of the sphere and also is the half of its diameter d:


r=\frac{d}{2}     (2)


Now, if we know the diameter of the black marble is 3cm, its radius is:


r=\frac{3}{2}cm     (3)


Substituting this value on equation (1):


V_{sphere}=\frac{4}{3}\pi{(\frac{3}{2}cm)}^{3}


Simplifying:

V_{sphere}=\frac{9}{2}\pi{cm}^{3}


V_{sphere}=14.137{cm}^{3}>>>>>This is an approximation of the volume of the marble


Note that 1{cm}^{3}=1ml, therefore the result above can be also written as 14.137ml


2. Answer: 64.137 ml


According to the Archimedes’ Principle a body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.


In this case, if we have a graduated cylinder with the capacity to contain 100 ml of water, and we fill it with 50 ml of water (as shown in the image attached) and then we add the black marble until it sinks; the water level will increase according to the principle explained above.

As the marble does not absorb water, the space it occupies displaces the water upwards and, in this way, it is possible to determine its volume or the final volume the water level indicates in the cylinder.


We already know the initial volume of water V_i in the graduated cylinder, which is 50 ml, and we know the volume of the marble V_m because we calculated it above. If we want to know the final volume of water level V_f we have to use the following relation:


V_{f}-V_{i}=V_{m}     (4)


and find V_f:


V_{f}=V_{i}+V_{m}     (5)


V_{f}=50ml+14,137ml    


Finally:

V_{f}=64,137ml>>>>>This is the final volume of the water level indicated in the graduated cylinder




Luden [163]3 years ago
7 0

the answer is 14cm

and 124mL

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Varvara68 [4.7K]

Answer:

Option D. Weight varies with location, but mass does not.

Explanation:

To know which option is correct, it is important that we have a background knowledge of mass and weight.

A brief summary of the difference between mass and weight is given below:

1. Mass is the quantity of matter present in an object while weight is the gravitational pull on an object.

2. The SI unit of mass is kilogram Kg) while that of weight is Newton (N)

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6 0
4 years ago
Can someone solve this problem and explain to me how you got it​
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1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

F=k\frac{q_1 q_2}{r^2}

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

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q_1' = 8q_1

q_2' = 8q_2

Substituting into the equation, we find the new force:

F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi
zzz [600]

Answer:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Explanation:

We know that the electric force equation is:

F=k\frac{q_{1}*q_{2}}{r^{2}}

  • k is the electric constant 9*10^{9} Nm^{2}/C^{2}
  • r is the distance between the particles
  • q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}

F_{31}=k\frac{q*2q}{r_{31}^{2}}

F_{31}=k\frac{2q^{2}}{r_{31}^{2}}

r(31) is the distance between 3 and 1

2. Now,  let's find the electric force between the third particle and the second particle.

F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}

F_{32}=k\frac{q*(-q)}{r_{32}^{2}}

F_{32}=-k\frac{q^{2}}{r_{32}^{2}}

r(32) is the distance between 3 and 2.

Now, r_{31}+r_{32}=2a or r_{32}=2a-r_{31}

The net force must be zero so:

F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex]   [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}

We just need to solve it for r(31)

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Therefore the distance from the origin will be:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

I hope it helps you!        

                 

 

     

4 0
4 years ago
ad for the distance from the sun tothe earth is 1.5 x 10"m. how long does it take for light from ths sun to reach the eath? give
Alexandra [31]

Answer:

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Explanation:

Use the formula for velocity as the distance covered by the light divided the time it takes: [tex]velocity=\frac{distance}{time}[/tex]

Use the information about the speed of light in vacuum: 300000000 \frac{m}{s} = 3*10^{8} \frac{m}{s}

and the information you are given regarding the distance between Sun and Earth: 1.5 * 10^{11} m

to solve the first velocity equation for the unknown time "t":

velocity=\frac{distance}{time} \\3*10^8\frac{m}{s} =\frac{1.5*10^11 m}{t} \\t=\frac{1.5*10^11 }{3*10^8} s= 500 s

we can convert second into minutes by dividing by 60: 500 s = 500/60 minutes = 8.3333... minutes

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puteri [66]

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

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