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2 years ago

HOW DO PROTONS, NEUTRONS AND ELECTRONS DIFFER?

Physics

2 answers:

aliina [53]2 years ago

6 0

Protons are positively charged and neutrons are neutral whereas electrons are negatively charged.

ahrayia [7]2 years ago

5 0

The protons has the positive charge, neutron has no charges and electrons have the negative charge.

Explanation:

The basic difference between the particles of the atom are their charges and their orientation in the atom. The proton and the neutron are present in the nucleus of the atom whereas the electrons are at the periphery of the atom. The nucleus is stationary in orientation having proton and neutron in it, whereas the periphery as the electrons are moving. Electrons revolves around the nucleus.

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Which one of the following is not a balanced equation

KATRIN_1 [288]

Answer:

Explanation:

i am not to sure!?!?!

sorry hopes this helps

5 0

2 years ago

Built an atom of Krypton which has an atomic number of 36?

gogolik [260]

Answer:

what's the question exactly because what you said was true but what's the question

7 0

2 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = 181 kg-m/s

relative to the dock. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = 0.12 m/s , away from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction toward the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = 0.62 m/s away from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from that at 0.62 m/s, and the person standing
on the dock sees you start to move away from him at 0.12 m/s, and that's the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

2 years ago

4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o

goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

(i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0

2 years ago

Wнιcн deмonѕтraтeѕ condυcтιon

jekas [21]

C.) cool feet walking across a hot pavement.

The reason because the other ones deals with radiation. Only C.) is the right answer because the feet is touching the hot pavement which is conduction.

3 0

2 years ago