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artcher [175]
1 year ago
11

Why is the following situation impossible? Two identical dust particles of mass 1.00 µg are floating in empty space, far from an

y external sources of large gravitational or electric fields, and at rest with respect to each other. Both particles carry electric charges that are identical in magnitude and sign. The gravitational and electric forces between the particles happen to have the same magnitude, so each particle experiences zero net force and the distance between the particles remains constant.
Physics
1 answer:
Igoryamba1 year ago
4 0

Answer:

This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:

a=\frac{F}{m}

Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.

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Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
What principles of science (like facts, laws, and theories) might help explain why similar investigations conducted in many part
natulia [17]

Answer:

Reproducibility of research

Explanation:

The principle of science that explains why similar experimental investigations conducted in different parts of the world could result in the same outcome is referred to as reproducibility.

<em>A good research or experiment in science must be reproducible, otherwise, the outcome of such an experiment might become inadmissible within the scientific community. It is a core principle of the scientific method that similar results should be obtained when an experiment or observational study conducted in one place is repeated in another place with the same procedure. Hence, an experiment must be reproducible in science in order for the outcome of such an experiment to be part of the general scientific knowledge. </em>

7 0
1 year ago
An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of
satela [25.4K]
<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>
8 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

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5 0
1 year ago
Read 2 more answers
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
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