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RoseWind [281]
3 years ago
7

What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the

distance between them is 1.0 x 10-12 m?
Physics
1 answer:
Agata [3.3K]3 years ago
3 0
The charge of the copper nucleus is 29 times the charge of one proton:
Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C
the charge of the electron is
e=-1.6 \cdot 10^{-19}C
and their separation is
r=1.0 \cdot 10^{-12} m

The magnitude of the electrostatic force between them is given by:
F=k_e  \frac{Qe}{r^2}
where k_e is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N
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The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a bure
Lapatulllka [165]

Answer:

γ = 0.06563 N / m

9.78% difference

Explanation:

Solution:-

- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.

- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.

                           γ = F / L

Where,

              F: Force imparted

              L: The length over which force is felt

- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:

                         m = M / n

                         m = 3.78 / 100

                        m = 0.0378 g        

- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:

                        F = m*g

                        F = 0.0378*9.81*10^-3

                        F = 0.000370818 N      

- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:

                        L = π*d

                        L = π*( 0.0018 )

                        L = 0.00565 m

- Then the surface tension would be:

                        γ = F / L

                        γ = 0.000370818 / 0.00565

                        γ = 0.06563 N / m

- The tabulated value of water's surface tension is given as follows:

                       γa = 0.07275 N/m

- We will determine the percentage difference between the value evaluated  and tabulated value as follows:

                     p.diff = \frac{gamma_a - gamma}{gamma_a} * 100\\\\p.diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\p.diff = 9.78 %

- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.

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