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2 months ago
11

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

ubatomic particles. Do you agree with her? Why or why not?
Physics
2 answers:
kicyunya [3.2K]2 months ago
4 0
No, that statement isn't correct because an atom is made up of two main regions and three subatomic particles: protons, neutrons, and electrons. Each of these particles is smaller than the atom itself, qualifying them as subatomic particles. To be classified as an atom, both protons and electrons are essential — for example, hydrogen has one proton and one electron. Neutrons don't influence the atomic electrical charge; they only add to the atomic mass.
Yuliya22 [3.3K]2 months ago
3 0

Answer:

An atom consists of two principal components: the nucleus and the electron cloud. Additionally, it contains three subatomic particles: protons, neutrons, and electrons. Therefore, I disagree with the claim that there are three main parts and only two subatomic particles.

Explanation:

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A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Yuliya22 [3333]

Answer:

The horizontal distance d that the ball covers before it lands is 1.72 m.

Explanation:

Given,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

We need to determine the horizontal distance d the ball travels before landing.

We need to calculate the time

Using the equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

Next, we can find the ball's velocity

Using the kinetic energy formula

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

By applying the conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

We substitute the values into the equation

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

Next, we determine the horizontal distance d the ball travels before landing

Using the distance formula

d =vt

Where. d = distance

t = time

v = velocity

We substitute the values into the formula

d=2.922\times 0.587

d=1.72\ m

Thus, the horizontal distance d that the ball travels before it lands is 1.72 m.

8 0
15 days ago
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
Softa [3030]

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

3 0
1 month ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
Softa [3030]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change (\Delta h)=-50 m

initial speed (v_1)=15 m/s

Final Speed (v_2)=30 m/s

Specific internal Energy (\Delta U)=-5 kJ/kg

according to the first Law of thermodynamics

Q-W=\Delta H

\Delta H=\Delta KE+ \Delta PE +\Delta U

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U

Q=W+\Delta KE+ \Delta PE +\Delta U

Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0
2 months ago
The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet
Keith_Richards [3271]
The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.
7 0
1 month ago
A supplier wants to make a profit by buying metal by weight at one altitude and selling it at the same price per pound weight at
Ostrovityanka [3204]

The appropriate choice is C.

In physics, the law of gravity helps us understand how gravity varies with height. As altitude increases, so too does the experience of gravity. Changes in altitude also result in variations in weight, though these differences are not particularly significant. Consequently, weighing metals at different heights shows negligible variance as the impact of gravity remains constant across them.

3 0
21 day ago
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