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Andrews
3 months ago
15

Suppose you sketch a model of an atom using the ones here as a guide. How would you build a model that is ionized? How would you

build a model that is radioactive?
Physics
1 answer:
Maru [3.3K]3 months ago
3 0

Answer:

Explanation:

An atom comprises three distinct particles: electrons, protons, and neutrons.

These particles vary in mass and electric charge, contributing to the atom's characteristics.

Electrons bear a negative charge, protons are positively charged, and neutrons possess no charge. An atom is electrically neutral when it has an equal number of electrons and protons, but this can change if particles are removed.

1: Ionized atom model - an ionized atom carries a net charge, which can be positive or negative.

To illustrate an ionized atom, one would need to reduce the number of either electrons or protons.

2: Radioactive atom model: A radioactive atom is characterized as unstable and retains excess energy in its nucleus, often due to added neutrons or protons.

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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
ValentinkaMS [3465]

Answer:

335°C

Explanation:

Heat that is either absorbed or released can be expressed as:

q = m C ΔT

where m represents mass, C signifies specific heat capacity, and ΔT indicates temperature change.

The heat gained by water equals the heat lost by copper.

mw Cw ΔTw = mc Cc ΔTc

The final temperature for both water and copper is the same, thus:

mw Cw (T - Tw) = mc Cc (Tc - T)

Provided:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Required: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
3 months ago
You are an investigator at the scene of a car crash. A Volvo of has hit an Alpha Romeo from the side and the two are inter-tangl
serg [3582]

Answer:

Let the Volvo's speed upon braking be v, and its mass be m1.

The speed of the Volvo just before the crash is represented as v1.

After applying the brakes, the skidding distance is noted as 30 m.

The equation governing this scenario is v1^2- v^2 = - 2 * a * s, where a= represents deceleration and s= denotes skidding distance.

Assuming the deceleration to be -0.3 g = - 2.94 m/s^2 (as provided by the Volvo website for the car's emergency system).

The equation can be modified to solve for v: v^2 = v1^2 + (2 * a * s )= v1^2 + 176.4.

After the collision, both vehicles merge and proceed together (m1+m2) at an initial speed of V (let's say).

They then move 12.25 m and eventually come to a stop.

Using the equation (0)^2 -(V)^2 = - 2 * a' * s', where a' signifies the combined deceleration and s' denotes the distance traveled by the two cars.

Let's assume the combined system’s deceleration is notably higher at (- g).

Utilizing the format V^2 = 2 * 9.8 * 12.25.

This results in V = 15.5 m/s.

Using momentum conservation in the X direction (East): m1 * v1 = (m1+ m2) * V * cos 15.

Solving for v1 gives: v1 = (1650+ 975) Kg * 15.5 m/s * cos 15 /1650 Kg = 23.8 m/s.

Hence, v^2 = V1^2 + 176.4 = (23.8)^2 + 176.4.

This results in v = 27.3 m/s = 61.1 mph (indicating that he exceeded the speed limit when applying the brakes).

In the Y-direction momentum conservation shows: m2 * v2 = (m1+m2) * V * sin 15.

Solving for v2 results in: v2 = (m1+m2) * V * sin 15/m2.

We conclude with v2 = 41.7 m/s (93.3 m/hr) indicating he also exceeded the speed limit.

5 0
2 months ago
Read 2 more answers
Add a third force that will cause the object to remain at rest. Label the new force F⃗ 3. Draw the vector starting at the black
kicyunya [3294]
The new force F3 is added in the same direction as F2. To analyze the forces acting on an object in this scenario, we observe that they operate along the vertical axis, with F1 acting upward and F2 downward. To determine the necessary vector F3 to counteract the net force, it's important to calculate the length difference between F1 and F2. The direction of F3 will match that of the smaller force. If F2 is less than F1, F3 will align with F2.
4 0
2 months ago
If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
Maru [3345]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Consequently, the temperature difference across the material will be \Delta T = 375 K

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

\Delta T is the temperature difference we need to determine

represents the thickness of the material\Delta x=2.5 cm =0.025 m

To isolate \Delta T from equation (1), we obtain:

\Delta T =\frac{Q \Delta x}{Ak}

Initially, we convert 3KW to W, resulting in:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

With all variables accounted for, we can substitute and calculate:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Thus, the temperature difference across the material will be \Delta T = 375 K

5 0
2 months ago
An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista
Sav [3153]

Answer:

        h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

  • when the object descends, its weight matches the spring's force

        weight = spring force

         mg = ky... equation 1

  • potential energy stored in a stretched spring = work done by the spring

        mgh = 0.5 x k x h^{2}....equation 2

  • Substituting from equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

                y =  0.5 x h

                2y = h

  • where y is 6.4, yielding the maximum elongation as

          h = 2 x 6.4 = 12.8 cm

6 0
3 months ago
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