Which statementsabout a neutral atom are correct? Check all that apply.1. A neutral atom is composed of bothpositively and negatively charged particles.
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.

Answers

Answer 1

Answer:

1, 2, 4, 5 are correct

Explanation:

1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).

2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.

3) This is not true because the positively charged particles in the nucleus are called protons.

4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.

5) This is true because opposite charges attract each other. And electron is a negative charge.

6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.

Related Questions

A gate with a circular cross section is held closed by a lever 1 m long attached to a buoyant cylinder. The cylinder is 25 cm in diameter and weighs 200 N. The gate is attached to a horizontal shaft so it can pivot about its center. The liquid is water. The chain and lever attached to the gate have negligible weight. Find the length of the chain such that the gate is just on the verge of opening when the water depth above the gate hinge is 10 m.
Which statements explain why theories change over time? Check all that apply. Scientists change the definition of theory to have their ideas accepted. New information and technology may be developed that influence the theory. Theories change with each new generation of scientists. New experimental methods provide new information. Theories may or may not be supported by new information.
I'm a little bit unsure about this question.
In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with radius 5.0 m. The ride moves at a constant speed and makes one complete circle in a time T=4.0s. What is the acceleration of the passengers? If the ride increases in speed so that T=3.0s, what is arad? (This question can be answered by using proportional reasoning, without much arithmetic.)
Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )x = vi(cos )tx = aytx = vxt (RIGHT ANSWER)

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answers

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=(GMm)/(r^2)$

Where $G=6.67*10^(-11)Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=(GMm)/(r^2)$

Which means:

$a=(GM)/(r^2)$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=(2GMd)/(r^2)$

$v=\sqrt{(2GMd)/(r^2)}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5*10^3m$ . With our values then we have:

$v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s$

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.

Answers

Answer:

-4.40

Explanation:

explanation is in attachment

How many meters will a car travel if its speed is 45 m/s in an interval of 11 seconds?Question 2 options:

A) 450 meters

B) 495 meters

C) 4.09 meters

D) 498 meters

Answers

Data given:

V=45m/s

t=11s

Δx=?

Formula needed:

V=Δx/t

Solution:

Δx=v×t

Δx=45m/s×11s

Δx=495m

According to my solution B) is the most accurate

495 is the answer !!

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

Explanation:

The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow

putting the values

π x .2945² x 12.1

= 3.3 m³ /s

To know the pipe's diameter at the refinery we shall apply the following formula

a₁ v₁ = a₂ v₂

a₁ v₁ and a₂ v₂ are volume rate of flow of liquid which will be constant .

3.3 = a₂ x 6.29

a₂ = .5246 m³

π x r² = .5246

r = .4087 m

= 40.87 cm

diameter

= 81.74 cm

When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Starting from rest, it takes two revolutions for the disk to reach full speed. Assuming that the angular acceleration of the disk is constant, what is its angular acceleration while it is speeding up

Answers

Final answer:

The angular acceleration of the disk drive in an old computer game system while speeding up is 1256 rad/s². This is calculated using kinematics in rotational motion, given the information on rotations, revolution time, and start from rest.

Explanation:

To calculate the angular acceleration of a disk drive in an old computer game system, we must use the concept of kinematics in rotation. When it is stated that it takes two revolutions to reach full speed, this implies that the total angular displacement is 4π radians (since one full revolution is 2π radians).

Given that the disk drive revolves once every 0.050 seconds, the final angular speed (ω) can be computed as 2π rad/0.050 s = 125.6 rad/s. Since the disk starts from rest, the initial angular speed (ω0) is 0. As a result, the total time taken (t) to reach full speed is 2*0.050s = 0.1 s.

We can then use the equation of motion in rotational form, α = (ω - ω0)/t, to calculate the angular acceleration. Hence the angular acceleration (α) is (125.6 rad/s - 0 rad/s) / 0.1 s = 1256 rad/s². Therefore, the angular acceleration of the disk drive is 1256 rad/s² while it is speeding up.

Learn more about Angular Acceleration here:

brainly.com/question/30238727

#SPJ12

Final answer:

The angular acceleration of the disk drive while it is speeding up is 8π rad/s².

Explanation:

The angular acceleration of the disk drive while it is speeding up can be determined by using the formula: angular acceleration = (final angular velocity - initial angular velocity) / time taken. In this case, the initial angular velocity is 0 (since the disk starts from rest) and the final angular velocity is 2 revolutions per 0.050 seconds. To convert revolutions to radians, multiply by 2π. The time taken is the time for two revolutions, so it is 2 * 0.050 seconds. Plugging in these values in the formula, we get:

Angular acceleration = (2 * 2π rad/s - 0) / (2 * 0.050 s) = 8π rad/s²

Learn more about Angular acceleration here:

brainly.com/question/32293403

#SPJ2

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.