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Gelneren [198K]

2 years ago

8

A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plan

e, acts on the particle. What is the torque about the origin on the particle?

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

25 Nm

Explanation:

Parameters given

Distance of the particle from the origin, r = 2.0 m

Force acting on particle, F = 25 N

Angle of force, θ = 30°

The torque acting on a particle is given as:

τ = r * F * sinθ

Where r = radius of axis of rotation. This is the same as the position of the particle on the x axis.

Therefore, the torque will be:

τ = 2 * 25 * sin30

τ = 25 Nm

The torque acting on the particle is 25 Nm.

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A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

2 years ago

Which of the following statements is/are true? Check all that apply. Check all that apply. The work done by a nonconservative fo

navik [9.2K]

Answer:

The work done by a nonconservative force depends on the path taken. <u>TRUE</u>

A nonconservative force permits a two-way conversion between kinetic and potential energies. <u>TRUE</u>

A potential energy function can be specified for a nonconservative force.

<u>FALSE</u>

The work done by a conservative force depends on the path taken.

<u>FALSE</u>

A conservative force permits a two-way conversion between kinetic and potential energies.

<u>FALSE</u>

A potential energy function can be specified for a conservative force.

<u>TRUE</u>

Explanation:

The work done by a nonconservative force depends on the path taken. <u>TRUE</u>

This kind of force can not be obtained from potential function so the work made by this kind of force depend of the path taken.

A nonconservative force permits a two-way conversion between kinetic and potential energies. <u>TRUE</u>

A nonconservative force permits conversion to kinetic energy plus potential energy during it made work over the system.

<u></u>

A potential energy function can be specified for a nonconservative force.

<u>FALSE</u>

Considering that the work made by kind of force depend of the taken path they can not be determined by a potential fuction.

The work done by a conservative force depends on the path taken.

<u>FALSE</u>

This asseveration is related with the previous one. So the conservative force can be deducted from a potential function thus their y work made do not depend of the path taken.

A conservative force permits a two-way conversion between kinetic and potential energies.

<u>FALSE</u>

This means that as conservative force is related from a potential function it only can modify the potential energy of the system.

<u></u>

A potential energy function can be specified for a conservative force.

<u>TRUE</u>

<u></u>

This is as consequence of the definition of conservative force that it can be determined from a potential function.

4 0

1 year ago

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across

enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0

1 year ago

Read 2 more answers

A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C

miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

= 1. 579 x 10 ^-22N

6 0

1 year ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

2 years ago