Question 4 (5 points)
What's the general formula for an alkene if n is the number of carbon atoms

From the darkest part of the object's shadow, you would be able to see a small amount of the light source. There would be a small amount of light that is visible, but it would be faint. On the other hand, from the lightest part of the shadow, you would be able to see much more of the light source. The light source would be far brighter and more visible, and you would be able to identify the source of the light.

24 volts is the voltage drop across each of the resistors in the parallel configuration.

When resistors are connected in parallel, they share the same voltage across them. Therefore, the voltage drop across each resistor in this scenario would be the same.

Given:

Power supply voltage (V) = 12 V

Resistance of each resistor (R) = 30 Ω

Since the resistors are in parallel, the total resistance (R_total) can be calculated using the formula:

1/R_total = 1/R1 + 1/R2

Substituting the values:

1/R_total = 1/30 Ω + 1/30 Ω

1/R_total = 2/30 Ω

R_total = 15 Ω

Now, we can find the current flowing through the resistors (I) using Ohm's Law:

I = V / R_total

I = 12 V / 15 Ω

I = 0.8 A

Since the voltage drop across each resistor is the same, we can find it using Ohm's Law:

V_drop = I * R

V_drop = 0.8 A * 30 Ω

V_drop = 24 V

Therefore, the voltage drop across each of the resistors in the parallel configuration is 24 volts.

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Answer:

Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.

Explanation:

The evaluation of the capacitor (in series) network is as follows;

Part A

Q = 3.2 μC

Part B

Q₁ = 3.2 μC

Part C

Q₂ = 3.2 μC

Part D

U = 76.8 μJ

Part E

U₁ = 34 2/15 μJ

Part F

U₂ = 53 1/3 μJ

Part G

V₁ = 21 1/3 V

Part H

V₂ = 26 2/3 V

A capacitor consists of pairs of conductors separated by insulators. Capacitors are used to store electric charge.

The specified parameters are;

The voltage across ab = 48 V

The capacitance of the first capacitor, C₁ = 150 nF

Capacitance of the second capacitor, C₂ = 120 nF

Part A

The total charge in a capacitor network can be found as follows;

\(C_{eq} = \left(\dfrac{1}{150} + \dfrac{1}{120} \right)^{-1} nF = \left(\dfrac{3}{200} \right)^{-1}nF\)

\(C_{eq} =\left(\dfrac{3}{200} \right)^{-1}nF=66\frac{2}{3} \, nF\)

\(Q_{eq} = C_{eq}\times V_{ab}\)

Therefore;

\(Q_{eq}\) = 66 2/3 nF × 48 V = 3,200 × 10⁻⁹ C = 3.2 μC

Part B

The charge in the 150 nF capacitor is obtained from the formula for the charge in a capacitor; Q = C × V as follows;

Q = C₁V₁ = C₂V₂

The charge in the capacitors, C₁ and C₂ are the same as the total charge of 3.2 μC

The charge, Q₁ on the 150 nF capacitor, C₁ is therefore, 3.2 nC

Part C

The capacitors, C₁ and C₂ are in series, therefore, the charge in each capacitor is equivalent to the charge in the circuit, which is 3.2 μC.

Therefore, the charge, Q₂, in the 120 nF capacitor, C₂ is 3.2 μC

Q₂ = 3.2 μF

Part D

The total energy stored in the network can be obtained using the formula;

U = (1/2)·C·V²

Where;

U = The energy in the capacitor

C = The equivalent capacitance of the network = 66 2/3 nF

V = The voltage

Therefore;

\(U = \dfrac{1}{2} \times C_{eq}\times V^2\)

\(U = \dfrac{1}{2} \times 66\frac{2}{3} \times 10^{-9}\times 48^2 = 76.8\)

Part E

The energy stored in the 150 nF capacitor is found as follows;

\(Q_{eq}\) = Q₁ = C₁ × V₁

V₁ = \(Q_{eq}\) ÷ C₁

Therefore;

V₁ = 3.2 μC ÷ 150 nF = \(21\frac{1}{3}\) V

U₁ = 0.5×C₁×V₁²

Part F

The energy stored in the 120 nF capacitor, U₂, can be found as follows;

V₂ = 3.2 μC ÷ 120 nF = \(26\frac{2}{3}\) V

U₂ = 0.5 × 150 nF × \(\left(26\frac{2}{3} \, V\right)^2\) = \(53\frac{1}{3}\, \mathrm{ \mu J}\)

Part G;

The potential difference across the 150 nF, obtained in Part E, is 21 1/3 V

Part H

The potential difference across the 120 nF, obtained in part F, is 26 2/3 V

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Divorce is viewed negatively by functionalists from a macro perspective. A functionalist would blame divorce on the breakdown of social structures rather than looking at the persons involved. Some of the structures that  have evolved over the years leading to the rise in divorce rates according to Functionalists is the increasing erosion of the traditional role of men and women in the society.

According to conflict theorists, divorce is the consequence of a disagreement over resources in a marriage.

Conflict theory holds that the disintegration of marriage occurs from rivalry for resources and authority within the marriage, just as it does in a society when groups compete for limited resources.

Choices, according to symbolic interactionists, are based on learnt behavior.

Symbolic interactionists believe divorce is the consequence of two persons affected by their friends and family members.

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Answer:

a) cost = $ 0.29 , b) cost = $ 0.046 ,  c) cost = $ 2.80

Explanation:

a) Let's find the energy consumed by this bulb, which is the product of the power and time

         energy = P t

         energy = 100 24 = 2400 W h

as the cost of energy is given in kWh we reduce this value

         Energy = 2400 Wh (1 kW / 1000W)

         Energy = 2.40 kWh

Calculate the cost of energy

           cost = 0.12 2.40

           cost = $ 0.288

           cost = $ 0.29

b) energy = 16.0  24 = 384 Wh

we reduce to kWh

      energy = 0.384 kWh

      cost = 0.12 0.384

      cost = $ 0.0461

      cost = $ 0.046

c) let's find the power consumed by the oven

       P = I V

       P = 20.0 220

       P = 4400 W

we look for the energy consumed

       energy = 4400  5.30

       energy = 23320 Wh

we reduce to kWh

       energy = 23.32 kWh

we calculate the cost of this energy

       cost = 0.12 23.32

       cost = $ 2,798

       cost = $ 2.80

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

the Given

*The number of mol of neon is n = 1 mol.

*The temperature of the neon gas is T = 288 K.

*The mass of the neon gas is m = 0.01 kg.

*The gas constant R is R = 8.31 J/(mol.K).

To find: The average velocity of atoms of neon (v).

According to the given equation,

Substitute the known values.

The average velocity of atoms of neon is 847.33 m/s. Thus, option (A) is the correct answer.

I am pretty sure that it's the oven

Answer: 2.72

Explanation: if this is not correct I don't know what to tell you other than don't use for answers and instead LEARN

Answer:

Isometric exercises are static and great for people who suffer from joint pain. Isometric exercises, like planks, don't require you to move or bend any joints. Isotonic exercises, like squats, involve straining the muscles while moving the joints and applying a constant amount of weight.

Explanation:

Answer:

Isometric exercises are static and great for people who suffer from joint pain. Isometric exercises, like planks, don't require you to move or bend any joints. Isotonic exercises, like squats, involve straining the muscles while moving the joints and applying a constant amount of weight.

Explanation:

Answer:

Explanation:

The momentum of an object can be found by using the formula

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

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The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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The volume of the parallelepiped is 83 cubic units.

The volume of a parallelepiped with sides give as a = ( 7,2 ,4 ) , b = ( 4,7 ,6 ) and c = ( 3,4 ,7 ).

The volume of a parallelepiped with adjacent sides a, b, and c is given by the scalar triple product (a × b) · c.

First, need to calculate the cross product of vectors a and b

a × b =

\(\left[\begin{array}{ccc}i &j&k\\7&2&4\\4&7&6\end{array}\right]\)

= (2 × 6 - 4 × 7) i - (7 × 6 - 4 × 4) j + (7 × 7 - 2 × 4) k

= -8 i - 26 j + 45 k

Now, calculate the scalar triple product

(a × b) · c = (-8)(3) + (-26)(4) + (45)(7) = 83

Therefore, the volume of the parallelepiped is 83 cubic units.

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For an airplane whose speed is 60 m/s is flying at an altitude of 500 m over the ocean toward a stationary sinking ship, the horizontal distance  is mathematically given as

x=606m

Generally, the equation for the y axis distance  is mathematically given as

y=vt+0.5gt^2

Therefore

500=0t+1.2(9.8)t^2

t=10.1s

In conclusion,  the horizontal distance

x=vt

x=60*10.1

x=606m

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Answer:

3, 5, and 6, that should be the answer

Answer:

B. 200 kgm/s

Explanation:

The initial momentum of the blue train can be calculated using the formula:

p = mv

where p is the momentum, m is the mass, and v is the velocity.

The mass of the blue train is 50 kg and its velocity is 4 m/s. Therefore, the initial momentum of the blue train is:

p = 50 kg x 4 m/s = 200 kgm/s

Therefore, the initial momentum of the blue train is 200 kgm/s, which is option B.

Answer:

D)

The mass of the reactants is greater than the mass of the products.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

\(5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ]\)

= 720000 [k]

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

Answer:

Halved

Explanation:

F=ma

Let case 1 (original) be:

\(F_{1}=m_{1} a_{1} \\\)

Case 2 (new) be:

\(F_{2}=m_{2} a_{2}\)

Mass is double:

\(m_{2}= 2m_{1}\)

Force kept the same:

\(F_{1} =F_{2}\)

Combine the equation and gives:

\(\frac{F_{1} }{F_{2}} =\frac{m_{1} a_{1} }{m_{2}a_{2} }\\\frac{F_{1} }{F_{1}} =\frac{m_{1} a_{1} }{2m_{1}a_{2} }\\1=\frac{a_{1} }{2a_{2} }\\a_{2}=\frac{1}{2} a_{1}\)

Acceleration is halved

Answer:

The scale pushes up on Margie: 590 N

The floor pushes up on the scale: 628 N

Explanation:

The free-body diagram for Margie and the scale is

Then, the magnitude force is equal to the normal force for each case.

For the first case, we get:

Fn = 590 N

For the second case, we get:

Fn = 590 N + 38 N = 628 N

So, the scale pushes up Margie with a force of 590 N and the floor pushes up on the scale with a force of 628 N.

Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r   --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

F₁ = 4 F₀

Answer:

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force

Explanation:

A measure of energy expended in moving an object; most commonly, force times displacement. No work is done if the object does not move.

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Explanation:

c willl fall fast then a and b

The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

The time of motion of the girl is calculated as follows;

h = vt + ¹/₂gt²

where;

Substitute the given parameters and solve for time of motion;

50.8 = 0 + ¹/₂(9.8)t²

2(50.8) = 9.8t²

101.6 = 9.8t²

t² = 101.6/9.8

t² = 10.367

t = √10.367

t = 3.22 seconds

vf = vi + gt

where;

vi is the initial vertical velocity = 0

vf = 0 + 9.8(3.22)

vf = 31.56 m/s

Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

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Answer:

Option "C": "4.5 g"

Explanation:

N0 = 36 g, Let half-life is T.

t = 3 T, n is number of half lives = t / T = 3

By using the decay law of radioactivity

N / N0 = (1 / 2)^n

where

"N0" be the "initial amount"

"N" be the "amount left"

"n" be the "number of half-lives"

N / 36 = (1/2)^3

N / 36 = 1 / 8

N = 36 / 8 = 4.5 g

Coefficient of Kinetic Friction = 0.2

Mass of the crate = 1000 kg

Force applied = 2000 N

Normal force is the force applied on the object by the surface. It is equal and opposite to the force of gravity

So, we can say that Normal force = | Gravitational Force |

Normal Force = | mg |

Normal Force = 1000 * 9.8

Normal Force = 9800 N

We know that:

coefficient of Kinetic friction = Friction force / Normal force

replacing the known values

0.2 = Friction force / 9800

Friction Force=  0.2 * 9800

Friction Force = 1960 N

We know that a force of 2000 N is being applied on the crate in the Horizontal direction

Frictional force is always opposite to the horizontal force. So, we can say that:

Applied force - Friction Force = Net Force

replacing the variables

2000 - 1960 = Net force

Net Force = 40N

Therefore, a net force of 40N is being applied on the Crate

From newton's second law of motion:

F = ma

replacing the variables

40 = 1000 * a

a = 40/1000

a = 0.04 m/s²

Hence, the crate will have an acceleration of 0.04 m/s²