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3 years ago

If a force does not act parallel to the resulting displacement, what is the effect on the work done by the force?

1 answer:

8 0

Then only the component of the force that's parallel to the displacement
is used to calculate the work. The component of force that's perpendicular
to the displacement doesn't move through any distance at all, so its contribution
to the total work is zero.

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True or False

Anon25 [30]

Explanation:

The work of the electric force does not depend on the trajectory or the path taken. Regardless of trajectory a and b we have the following

Vide picture

$W_{ab} = \frac{K Q q}{R_{a} } \frac{K Q q}{R_{b} }$

Wab: Work

K: coulomb constant (9·109 N·m2/C2)

Ra,Rb: distances

Qq: electric charges

3 0

3 years ago

Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the freq

mel-nik [20]

Answer:

$\mathbf{f_o =1.87 \times 10^{11} \ Hz}$

Explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

$U_o = \dfrac{1}{2}kR_o^2$

Making K the subject;

$K = \dfrac{2 U_o}{R_o^2}$

$k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}$

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

$f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}$

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

$f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}$

$\mathbf{f_o =1.87 \times 10^{11} \ Hz}$

6 0

3 years ago

Why do so many of the people of India live in the Ganges River Valley?

irga5000 [103]

The river provides fertile soil for farming

5 0

3 years ago

Read 2 more answers

Find the orbital period (in years) of an asteroid whose average distance from the sun is 27

dmitriy555 [2]

Using Kepler's third law which is defined as the square of the average distance is directly proportional to the cube of the period. It is expressed as P^2 = a^3, Given that the a = average distance is given, the period would be much easier to compute. P = sqrt(27^3) = 140

7 0

3 years ago

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p

Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m$

The image distance is 0.051 m

When u = 50 cm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m$

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

3 0

3 years ago