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natta225 [31]
3 years ago
15

A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. what is the volume of the ball?

Physics
2 answers:
lesantik [10]3 years ago
6 0
Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3
densk [106]3 years ago
6 0

Answer:

Volume,V=4.82\ cm^3

Explanation:

Given that,

Mass of the lead ball, m = 55 grams

Density of lead ball, d=11.4\ g/cm^3

We need to find the volume of the ball. The formula of density is given by :

d=\dfrac{m}{V}

V=\dfrac{m}{d}

V=\dfrac{55\ g}{11.4\ g/cm^3}

V=4.82\ cm^3

So, the volume of the the lead ball is

V=4.82\ cm^3. Hence, this is the required solution.

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

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Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

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

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

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

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
Calcular la rapidez promedio
Vlada [557]

Answer:

  v_average = 15 m / s

Explanation:

The average speed can be found in two ways,

* taking the distance traveled and divide it by the time spent

* taking the velocities in each time interval and then finding the weighted average by the time fraction

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Let's apply this last equation

               

Total time is

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               t = 10 + 10 = 20   min

              v_average = 10/20 10 + 10/20 20

              v_average = 10/2 + 20/2

              v_average = 15 m / s

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