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3 years ago

Which particles in an atom could demonstrate that opposite charges attract?

Physics

2 answers:

Oxana [17]3 years ago

4 0

Answer:

Choice D. a proton and an electron.

Explanation:

Each proton carries a positive charge of $+1$ unit.

Each electron carries a negative charge of $-1$ unit.

As their name suggests, neutrons are neutral. They do not carry any electrostatic charge. Because of that, neutrons can't attract or repel neutrons or protons with electrostatic forces.

On the other hand, electrons and protons carry opposite charges. Therefore, an electron and a proton would attract each other electrostatically.

However, since protons are all positive, two protons would repel each other electrostatically.

kompoz [17]3 years ago

3 0

Answer:

The correct answer is D

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Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

6 0

3 years ago

A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wi

Maksim231197 [3]

Complete Question

A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop is

a. counterclockwise

b. clockwise

c. there is no current in the loop because magnetic field is uniform

d. there is no current in the loop because magnetic field does not change

Answer:

Option A is the correct answer

Explanation:

According to the question the loop of wire contracts to half it initial diameter and will mean that less number of electric field line will pass through the loop and this change in magnetic flux will cause current to flow in the loop of wire and from Lenz's law this current will in the opposite direction of what produced it which is the change in magnetic flux so the current will flow in a counterclockwise direction

6 0

3 years ago

Study the motion map shown. what do the circled vectors represent?

BabaBlast [244]

Answer:

It's option d - Negative acceleration

Explanation:

Let's start by demonstrate why it's not option b - Speed : Speed is a scalar quantity so it can not be represented by a vector

Let's check that the green vectors represent velocity (velocity is a vector quantity, velocity is a direction aware, while speed is just a scalar)

Now let's show that the circled vectors are acceleration vectors:

Mathematically position X , velocity V and acceleration A are:

$\frac{dX}{dt} = V$ and $\frac{dV}{dt} = A$

Where X, V, A are vectors and $\frac{d(a)}{dt} = b$ indicates the derivate a of a time is equal to b.

So, this show that acceleration is a rate respect of time of velocity ⇒ When acceleration is positive, velocity increments, when acceleration is negative, velocity decrements.

The above explanation correspond to the motion map shown, getting demonstrated that the answer is D - Negative acceleration

3 0

3 years ago

You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

Dimas [21]

Answer:

Explanation:

mass of baseball, m = 0.145 kg

initial velocity, u = 12 m/s upward

(a) final velocity, v = u / 2 = 6 m/s

Let the height is h.

Use third equation of motion

v² = u² - 2gh

6 x 6 = 12 x 12 - 2 x 9.8 x h

36 - 144 = - 19.6 x h

h = 5.51 m

(b) initial kinetic energy, K = 0.5 x m x u² = 0.5 x 0.145 x 12 x 12 = 10.44 J

Final kinetic energy, K' = K/2

0.5 x m x v² = 10.44 /2

0.5 x 0.145 x v² = 5.22

v = 8.5 m/s

v² = u² - 2gh

8.5 x 8.5 = 12 x 12 - 2 x 9.8 x h

72.25 - 144 = - 19.6 x h

h = 3.66 m

7 0

4 years ago

I need help with question 8 .

Vesnalui [34]

The bike is maintaining "constant velocity". He's moving at 15 m/s when we see him for the first time, 15 m/s later that day, and 15 m/s next week.

The car starts from zero, and goes 4.0 m/s FASTER each second. After one second, it's going 4.0 m/s. After 2 seconds, it's going 8 m/s. And after 3 seconds, it's going 12 m/s.

This is the point at which the question wants us to compare them ... 3 seconds. The bike is moving at 15 m/s and the car has sped up to 12 m/s. The bike is moving faster than the car.

If we hung around and kept watching for another second, the car would then be moving at 16 m/s, and would be moving faster than the bike. But we lost interest after answering the question, and we left at 3 seconds.

5 0

3 years ago