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Anestetic [448]
3 years ago
11

Seven little spheres of mercury, each with a diameter of 2 mm. When they coalesce to form a single sphere, how big will it be (i

.e. what is its diameter)? How does its surface area compare with the total surface area of the previous Seven little spheres?
Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

<em>The total surface area of the seven little spheres is 1.91 times the total surface area of the bigger sphere.</em>

Explanation:

<u>Volume of a Sphere</u>

The volume of a sphere of radius r is given by:

\displaystyle V=\frac{4}{3}\cdot \pi\cdot r^3

The volume of each little sphere is:

\displaystyle V_l=\frac{4}{3}\cdot \pi\cdot 2^3

V_l=33.51\ mm^3

When the seven little spheres coalesce, they form a single bigger sphere of volume:

V_b=7*V_l=234.57\ mm^3

Knowing the volume, we can find the radius rb by solving the formula for r:

\displaystyle V_b=\frac{4}{3}\cdot \pi\cdot r_b^3

Multiplying by 3:

3V_b=4\cdot \pi\cdot r_b^3

Dividing by 4π:

\displaystyle \frac{3V_b}{4\cdot \pi}= r_b^3

Taking the cubic root:

\displaystyle r_b=\sqrt[3]{\frac{3V_b}{4\cdot \pi}}

Substituting:

\displaystyle r_b=\sqrt[3]{\frac{3*234.57}{4\cdot \pi}}

r_b=3.83\ mm

The surface area of the seven little spheres is:

A_l=7*(4\pi r^2)=7*(4\pi 2^2)=351.86\ mm^2

The surface area of the bigger sphere is:

A_b=4\pi r_b^2=4\pi (3.83)^2=184.33\ mm^2

The ratio between them is:

\displaystyle \frac{351.86\ mm^2}{184.33\ mm^2}=1.91

The total surface area of the seven little spheres is 1.91 times the total surface area of the bigger sphere.

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2. An electron and a proton are separated by 5 cm:
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3 years ago
I will be so thankful if u answer correctly!!​
olga_2 [115]
<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force F_{r} is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma    -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F -  F_{r}

m = 50kg

a = 0   [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F -  F_{r} = m x a

F -  F_{r} = 50 x 0

F -  F_{r} = 0

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<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force F_{r} is opposing the movement of the box.

∑F = 1.5F -  F_{r}

m = 50kg

a =  0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F -  F_{r} = m x a

1.5F -  F_{r} = 50 x 0.1

1.5F -  F_{r} = 5            ---------------------(iii)

<em>Substitute </em>F_{r}<em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5            

0.5F = 5            

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

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Answer:

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