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4 years ago

Please help...........................

Physics

2 answers:

Marianna [84]4 years ago

6 0

Hi!

Neutrons are neutral, which means they don't exactly have an electrical charge. It's because of this neutral charge that it is represented with a '0'.

On the other hand, protons and electrons <em>do </em>have electrical charges. Electrons flow around the outside of the nucleus, with a negative charge.

Protons are stored in the nucleus with the neutrons, holding a positive charge.

Hopefully, this helps! =)

borishaifa [10]4 years ago

6 0

Here are some tips to help you remember the difference between protons, neutrons, and electrons.

Protons-- Think pro. Pro means positive (as in pros and cons) and the proton is often represented with the + sign. +, positive, pro, proton.

Neutrons-- Neutron sounds like neutral. Neutrally charged particles are often represented with an O. I personally think of Jimmy Neutron here. Every time his inventions went bad, his friends would say things like "Oh boy." O, Jimmy Neutron, neutron.

Electrons-- Think about how negative the United States are about the current president and how he was ELECTED. The - sign represents electrons. -, negative, presidential election, electron.

So the answer to your problems would be:
1. B-- electrons are negative
2. A-- protons are positive
3. C-- neutrons are neutral

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wlad13 [49]

Answer:

the correct answer is B

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Nitrogen oxides in car exhaust react with sunlight to produce

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Photochemical smog which creates a brown haze

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2 years ago

At one instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The car has a greater veloc

Helen [10]

Answer:

Explanation:

Acceleration is change in velocityΔv in respect to timeΔt

so if the velocity of the car is greater than the truck it does not mean that the car acceleration is greater than the truck.

Sometimes with constant velocity it means no accelaration ,but the truck may have accelaration

so, higher velocity of the car does not mean higher acceleration

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3 years ago

A yoyo with a mass of m = 150 g is released from rest as shown in the figure.

avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

I is moment of inertia
α is angular acceleration
T is tension in the rope
r is inner radius
R is outer radius
f is frictional force

rT - Rf = Iα ----- (1)

T - f = Ma -------- (2)

a = Rα

where;

a is the linear acceleration of the yoyo

Torque equation for frictional force;

$f = (\frac{r}{R} T) - (\frac{I}{R^2} )a$

solve (1) and (2)

$a = \frac{TR(R - r)}{I + MR^2}$

since the yoyo is pulled in vertical direction, T = mg $a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2$

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0

2 years ago

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 965m/s. the mass of the hunter (in

weqwewe [10]

When the gun is fired horizontally :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 m/s

M = mass of the hunter including gun = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (965) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.056 m/s

so recoil velocity comes out to be 0.056 m/s

When the gun is fired at angle 56.0⁰ above the horizontal :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 Cos56 = 539.62 m/s

M = mass of the hunter including gun = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (539.62) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.031 m/s

so recoil velocity comes out to be 0.031 m/s

3 0

3 years ago