1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Solnce55 [7]
3 years ago
7

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of

electrons? (b)What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?
Physics
1 answer:
Advocard [28]3 years ago
5 0

Answer:

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

Explanation:

(a) On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

The mass of an atom is given by the sum of the masses of the protons, neutrons and electrons. Electrons has lower mass than protons and neutrons, so they have a minor contribution to the total mass of the atom.    

When an object is electrically neutral it means that it has the same number of protons and electrons. For the case of an object positively charged, the rate of protons is greater than the number of electrons. That means that atom lose electrons so the mass will decrease in a very small factor.

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

For the case when the object is negatively charged, it means that the atom gains electrons from another object, leading to the conclusion that the mass of the atom will increase in a very small factor.  

Key values:

Electron mass: 9.1095×10⁻³¹ Kg

Proton mass: 1.67261×10⁻²⁷ Kg

Neutron mass: 1.67492×10⁻²⁷ Kg

You might be interested in
ASAP WILL GIVE BRAINLIEST Take a piece of paper, some nails and a magnet bar, place a magnet at the centre of the paper, and spr
Zanzabum

Answer: find the answer in the explanation

Explanation:

When a magnet is placed at the centre of the paper, and the nails are sprinkled on the paper, what will happen to the nails is that, the nails will form a pattern on the paper according to the magnetic field of the bar magnetic pole.

Other phenomena you can observe are:

1.) The nails will align themselves and show some lines of forces which is equivalent to the magnetic field lines

2.) The direction of the line of forces

3.) The strength of the magnetic field pole.

7 0
3 years ago
a heavy jar sits on top of a 3.4 m shelf with a gravitational potential energy of 180 j. What is the mass of the jar?
sp2606 [1]

Answer:

5.2941176471 kg or 5294.1 grams

Explanation:

g.p.e= mgh

g.p.e/gh=m

180j/10×3.4= m

180/34= m

5.2941 kg= m

6 0
2 years ago
A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain
Mama L [17]

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.  

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

4 0
3 years ago
Describe what the effect of increasing the power of a camera would have on the battery life
patriot [66]

Answer:

. Cut Down on the LCD

The biggest battery drain in a camera is the LCD – both the rear screen and the electronic viewfinder. This is the big reason why DSLRs almost always have longer battery life specifications than mirrorless cameras – the optical viewfinder lets you skip LCDs altogether.

However, if you use your DSLR in live view, you’ll notice that its battery life slides dramatically. Side by side against a mirrorless camera, there’s actually a good chance it will die first. LCDs just take a lot of power to run.

What does this imply? Quite simply, you should always do what you can to cut down on LCD usage when your battery is running low.

For DSLR users, that means switching to the optical viewfinder. For mirrorless photographers, it means turning off the camera frequently, or setting it so the viewfinder only activates when you hold it to your eye.

And regardless of the camera you use, drastically cut down on the amount of time you spend reviewing photos. Chimping has its place, but not while your battery warning is blinking red.Optimize Your Battery Saver Settings

Most cameras have menu options designed to improve battery life and maximize your shooting time. For example, the “metering timeout” setting lets you select how long you want the camera to wait during inactivity before shutting off its metering system.

Beyond that, a number of cameras today have an “Eco mode” that minimizes power consumption from the camera’s LCD. On the Canon EOS R, for example, Eco mode dims and then turns off the LCD when not in use, improving your battery life significantly – from 370 to 540 shots per charge, according to Canon’s official specifications.

It’s also important to note that mirrorless cameras are generally more efficient using the rear LCD than the electronic viewfinder. In terms of the EOS R again, Canon only rates 350 shots using the EVF, with no Eco mode to improve it. On the Sony side of things, the new A7R IV is rated for 530 shots via the viewfinder and 670 via the rear LCD.

If none of that applies to you, one option at your disposal is always to lower the brightness of your rear LCD. It might make photography a bit trickier in bright conditions, but the payoff is getting the shot rather than missing it completely due to a dead battery.

Other camera settings and extras that harm battery life include:

Image stabilization (both in-body and in-lens)

Popup flash

Bluetooth and WiFi

Most external accessories: GPS dongles, lightning triggers, wireless remote releases, shotgun mics, etc.

Sometimes, these capabilities are essential for your photo, so it’s worth the battery life sacrifice. But if you’re down to your last bar, double check to ensure that you’re not using any of the above settings or accessories without good reason.

8 0
3 years ago
The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of
Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is  =4.51m/s  

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

        Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

  Now to obtain the initial energy stored

               Let U denote the initial energy stored and

                      U  = \frac{1}{2} kx^2

Where x  is the length the spring is displaced

k is the force constant of the string

         U = \frac{1}{2} * 627 * (0.6)^2

          = 112.86 J

   Now referring to the formula above

i.e          Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

                \frac{1}{2} mv^2 = 112.86 - \mu_kmgx

                v^2 = \frac{2(112.86 -\mu_kmgx)}{m}

                v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}

and we are told that coefficient of friction  = 0.4 and the mass is 9 kg ,the acceleration due to gravity = 9.8m/s^2  this displacement length of spring = 0.6

  Therefore   v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }

                        =4.51m/s      

           

8 0
3 years ago
Other questions:
  • The soccer player is running at a speed of 4m/s when he slides on the ground, coming to a stop distance of 2m. What is the accel
    14·1 answer
  • The _______ effect draws a small character below the bottom of the text.
    9·1 answer
  • 2. Montesquieu's view of the separation of powers was later expressed in the United States Government through which document? A.
    11·1 answer
  • Which equals 23 kilograms?
    10·2 answers
  • when a body of mass 0.25kg is acted upon by a force, the velocity changes from 5m/s to 7.5m/s. Calculate the work done by the fo
    5·1 answer
  • A tank has a gate that automatically opens if the water levelhis high enough. The gate has a squarecross section of side1m and c
    11·1 answer
  • • 2. A 70 kg man on a 100 kg boat throws a ball. The boat moves backwards 5 meters in 10 seconds. What is
    8·1 answer
  • How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead
    11·1 answer
  • Which of the following phenomena can be used to support the particle theory of light? Select all that apply.
    5·1 answer
  • Choose one inner planet and one outer planet and compare their similarities and differences.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!