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beks73 [17]
3 years ago
15

There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres

are placed in contact then set at their original distance. Calculate the magnitude of the force between the two spheres at the final position. The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of N.
Physics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.

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Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

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v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

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Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

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s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

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This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

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3 years ago
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Answer:

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3 years ago
What is the voltage, V2, in units of Volts, across resistor R2 in the circuit shown below where VS = 4V, R1 = 14 Ohms and R2 = 3
svetlana [45]
<h2>Correct answer:</h2>

\boxed{v_{out}=2,85V}

<h2>Explanation:</h2>

We can use voltage divider to solve this problem that is defined as the passive linear circuit producing an output voltage v_{out} that is a fraction of its input voltage v_{in}. So we can use the formula:

v_{out}=\frac{R_{2}}{R_{1}+R_{2}}v_{in}, \ where \ v_{in}=v_{s}=4V \\ \\ \therefore v_{out}=\frac{35}{14+35}(4) \\ \\ \therefore v_{out}=2,85V

3 0
3 years ago
As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the dens
azamat

Answer:

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Answer:

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