Car A is accelerating in the direction of its motion at the rate of 3 ft /sec2. Car B is rounding a curve of 440-ft radius at a constant speed of 30 mi /hr. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 45 mi /hr for the positions represented.

The calculated heat Q is [tex]\( 1.22 \times 10^4 \mathrm{~J} \),[/tex] and the positive sign indicates that heat is being added to the system.

Let's break down the given answer step by step:

1. Given Values:

Temperature of gas, [tex]\( T = 22^{\circ} \mathrm{C} = 295 \mathrm{~K} \)[/tex]

Work done on the gas to compress it, [tex]\( W = -353 \mathrm{~J} \)[/tex](negative because work is done on the gas)

  Final temperature, [tex]\( T_f = 145^{\circ} \mathrm{C} = 418 \mathrm{~K} \)[/tex]

  Number of moles,[tex]\( n = 8.2 \) moles[/tex].

2. First Law of Thermodynamics:

The first law of thermodynamics is given by [tex]\( Q = \Delta U + W \),[/tex] where Q is the heat added to the system, [tex]\( \Delta U \)[/tex] is the change in internal energy, and W is the work done on the system.

For a monoatomic gas,[tex]\( C_v = \frac{3}{2} R \),[/tex]

where R is the universal gas constant [tex](\( R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \)). So, \( Q = n C_v \Delta T + W \).[/tex]

3. Calculation:

  Substituting the given values into the equation:

[tex]\[ \begin{aligned} Q & = \left(8.2 \times \frac{3}{2} \times 8.314 \times (418 - 295)\right) + (-353) \\ & = 12225 \mathrm{~J} = 1.22 \times 10^4 \mathrm{~J} \end{aligned} \][/tex]

4. Interpretation:

  Since Q is positive, it means heat is flowing inside the cylinder. A positive value of Q indicates that the system is gaining heat, and in this case, it's due to both the compression work done on the gas and the increase in internal energy.

In summary, the calculated heat Q is [tex]\( 1.22 \times 10^4 \mathrm{~J} \),[/tex] and the positive sign indicates that heat is being added to the system.

143μH

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l                 --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

From the question;

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

But;

A = π d² / 4                     [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

Substitute these values into equation (i) as follows;

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

The inductance, in microhenrys, of Tarik's solenoid should be considered as the 143μH.

Since

The inductance (L) of a coil wire (e.g solenoid) should be provided by

L = μ₀N²A / l                 --------------(i)

here,

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

So,

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

So,

A = π d² / 4              

= 3.142 x 0.00949² / 4

= 7.1 x 10⁻⁵m²

Now

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

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Answer:

10ns

Explanation:

Suppose the EM wave travels at light speed [tex]c = 3\times10^8 m/s[/tex]. A change in travel distance of the order of 3 m would result of a change in travel time of

[tex] \Delta t = \frac{\Delta s}{c} = \frac{3}{3\times10^8} = 10^{-8} s = 10 ns[/tex]

To detect a change in travel distance of about 3 m, a GPS receiver must measure a change in travel time of approximately 10 nanoseconds. This is calculated using the speed of light and the time = distance/speed formula.

To determine the associated travel time that must be measured for a change in travel distance of 3 meters, we need to know the speed of electromagnetic waves, which is the speed of light, typically denoted 'c'. The speed of light is approximately 3.00 x 108 meters per second (m/s).

Now we can calculate the time it takes for light to travel a given distance using the formula time (t) = distance (d) / speed (s). Substituting for our given distance (3 m) and the speed of light, we get t = 3 m / 3.00 x 108 m/s. This results in a time of approximately 1 x 10-8 seconds or 10 nanoseconds (ns).

Therefore, to detect a change in travel distance of about 3 m, the GPS receiver must be able to measure a change in travel time of roughly 10 nanoseconds.

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Answer with Explanation:

We are given that

Distance between two parallel long wires=r=3.3 cm=[tex]\frac{3.3}{100}=0.033m[/tex]

1 m=100 cm

[tex]\frac{F}{l}=5\times 10^{-5} N/m[/tex]

[tex]I_1=0.62 A[/tex]

a.We have to find the current in the second wire.

We know that

[tex]\frac{F}{l}=\frac{2\mu_0I_1I_2}{4\pi r}[/tex]

Using the formula

[tex]5\times 10^{-5}=\frac{2\times 10^{-7}\times 0.62\times I_2}{0.033}[/tex]

Where [tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]

[tex]I_2=\frac{5\times 10^{-5}\times 0.033}{2\times 10^{-7}\times 0.62}[/tex]

[tex]I_2=13.3 A[/tex]

Hence, the current in the second wire=13.3 A

b.We are given that the wires repel each other.When the current carrying in the wires in opposite direction then, the wires repel to each other.

Hence,the two  currents in opposite directions.

Answer:

The time is 0.713 sec.

Explanation:

Given that,

Weight of water = 4 liters

Initial temperature = 128°C

Power = 1400 Watts

Final temperature = 708°C

Weight = 1.1 kg

Specific heat of steel = 0.46 kJ/kg°C

Specific heat of water = 4.18 kJ/kg°C

We need to calculate the heat gained by bucket

Using formula of heat

[tex]Q_{b}=mc\Delta T[/tex]

Put the value into the formula

[tex]Q_{b}=1.1\times0.46\times(70-12)[/tex]

[tex]Q_{b}=29.348\ kJ[/tex]

We need to calculate the heat gained by water

Using formula of heat

[tex]Q_{w}=mc\Delta T[/tex]

Put the value into the formula

[tex]Q_{w}=4\times4.18\times(70-12)[/tex]

[tex]Q_{w}=969.76\ kJ[/tex]

We need to calculate the total heat

Using formula of heat

[tex]Q=Q_{b}+Q_{w}[/tex]

Put the value into the formula

[tex]Q=29.348+969.76[/tex]

[tex]Q=999.108\ kJ[/tex]

We need to calculate the time

Using formula of time

[tex]t=\dfrac{Q}{P}[/tex]

Put the value into the formula

[tex]t=\dfrac{999.108}{1400}[/tex]

[tex]t=0.713\ sec[/tex]

Hence, The time is 0.713 sec.

Answer:

The time is [tex]110.16\times10^{-3}\ sec[/tex]

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

[tex]i_{0}=\dfrac{V_{0}}{R}[/tex]

Put the value into the formula

[tex]i_{0}=\dfrac{150}{1.8\times10^{3}}[/tex]

[tex]i_{0}=83.3\times10^{-3}\ A[/tex]

We need to calculate the time

Using formula of current

[tex]i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}[/tex]

Put the value into the formula

[tex]50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}[/tex]

[tex]\dfrac{50}{83.3}=e^{\frac{-t}{RC}}[/tex]

[tex]\dfrac{-t}{RC}=ln(0.600)[/tex]

[tex]t=0.51\times1.8\times10^{3}\times120\times10^{-6}[/tex]

[tex]t=110.16\times10^{-3}\ sec[/tex]

Hence, The time is [tex]110.16\times10^{-3}\ sec[/tex]

The current across the student's chest will exceed the danger level for approximately 216 seconds.

The current that flows through the student's chest when he accidentally touches the terminals of the 120 mF capacitor can be calculated using Ohm's law: I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, the voltage is 150 V and the resistance of the student's body is 1.8 kΩ. Plugging in these values, we can find the current.

I = (150 V)/(1.8 kΩ) = 0.0833 A

To find how long the current will exceed the danger level of 50 mA, we need to calculate the time it takes for the charge on the capacitor to dissipate through the student's body. Since the current is constant, we can use the formula Q = I*t, where Q is the charge and t is the time. Rearranging the formula, we can solve for t.

t = Q/I = (120 mF * 150 V) / (0.0833 A) = 216 s

Therefore, the current will exceed the danger level of 50 mA for approximately 216 seconds.

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Answer:

58.6 N

Explanation:

We are given that

[tex]q_1=-7.97\mu C=-7.97\times 10^{-6} C[/tex]

[tex]q_2=3.55\mu C=3.55\times 10^{-6} C[/tex]

Using [tex]1\mu C=10^{-6} C[/tex]

[tex]r=6.59 cm=6.59\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

The magnitude of force that one particle exerts on the other

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Where [tex]k=9\times 10^9[/tex]

Substitute the values

[tex]F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}[/tex]

F=58.6 N

Answer: 3333333222135790075

Explanation:Set term u equal to initial velocity for simplicity

Set V equal to final velocity for simplicity

2

To begin this problem, one must look at the system to have multiple stages. These being before and after hitting the puck. In these first few steps, we look at BEFORE the human hits the puck

3

This collision is elastic because the puck and the human do not join together after interaction

4

Because the initial velocity of both the puck and the human are both 0, the terms on the left of the equal sign become 0

5

Solving for the final velocity of the human gives this formula. This number should be negative as the negative indicates the direction he is going (left)

The final velocity of the puck is already given in the problem

6

Because the ice is frictionless, the final velocity before hitting the puck is equal to the initial velocity after hitting the puck

Now we begin to look at the system AFTER the puck has been hit

7

Using the formula for final position allows us to solve for time it takes the puck to travel the distance given

8

9

Solve for time

10

We can now use the formula for the final position of the human to solve for the final answer

11

12nuewnfunw

Plugging in formulas from steps 5 and 9 gives the final answer

The distance traveled by the player ( recoil ) in the time the puck reaches the goal is 0.043m.

According to the law of conservation of momentum, the sum of the momentum of the object before and after the collision must be equal.

m₁ u₁ + m₂ u₂ =   m₁ v₁ + m₂ v₂

where m₁ and m₂ is the mass of the objects, u₁ and u₂ are initial speed while v₁ & v₂ is final speed.

Given the initial velocity of the player is u₁ = 0 and the puck is u₂ = 0

The mass of the player m₁ = 84 Kg

The mass of the puck, m₂ = 0.150 Kg

The final velocity of the puck, v₂ = 36 m/s

From the law of conservation of momentum, find the velocity of the player:

m₁ u₁ + m₂ u₂ =m₁ v₁ + m₂v₂

84 × 0 + m ×0 = 84 × v + 0.150 ×  36

v = - 0.064 m/s

A negative sign shows the player and puck moving in the opposite direction.

Now, we calculte the time taken for the puck to trach the goal:

Time = Distance/ Velocity

t = 24/36 = 0.667 sec

Next, we calculate the distance traveled by the player( recoil ) in the time of 0.667 seconds.

Distance =  Velocity× time

S = 0.064 ×0.667

S = 0.043m

Therefore, the distance covered by the player in the time the puck reaches the goal is 0.043m.

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(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

Answer:

108 Watts

Explanation:

The total circuit resistance when the resistors are connected in series is

               R + R + R = 3R

When he resistors are connected in parallel, the resistance reduces from 3R in the series circuit to become;

             [tex]\frac{1}{R} + \frac{1}{R} + \frac{1}{R}[/tex]

                   = [tex]\frac{R}{3}[/tex] Ω

[tex]Power = \frac{V^{2}}{R}[/tex]

The voltage supply was given to be constant for both the series and parallel circuits. This implies that V² is constant and power is inversely proportional to resistance.

Therefore;

Power for the parallel connected circuit = [tex]\frac{3R}{\frac{R}{3} } * 12 W[/tex]

                            = 9 × 12 W = 108 Watts

Answer:

Explanation:

Potential difference, V and let each resistance, R

Resistors are in series, total resistance, Rₓ = R1 + R2 + R3

= R + R + R

= 3R

Power, P = V²/Rₓ

12 = V²/3R

V²/R = 36

Resistors are in parallel, total resistance, 1/Rₓ = 1/R1 + 1/R2 + 1/R3

Rₓ = R/3

P = V²/Rₓ

P = V²/(R/3)

P = 3(V²/R)

= 3(36)

= 108 W.

Answer:

Pithball electroscope is used to determine if any object has static charge.

It consists of one or two small balls of a lightweight non-conductive substance. When this material is moved near an object having static charge, polarization will be induced in the atoms of pithballs which will either attract or repel the object depending on the nature of the charge in it.

It will not move in case it is brought near to a neutral object.

Similarly, the pith ball will move when it will brought into the magnetic field of the magnet as it will also induce polarization within the atoms of the pithballs.

Answer:

The electric force increases by a factor of 4.

Explanation:

The electric force between two charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance d can be calculated using Coulomb's Law:

[tex]F=\frac{kq_1q_2}{d^2}[/tex]

where [tex]k=9\times10^9Nm^2/C^2[/tex] is the Coulomb constant.

If the value of each charge is doubled, then we will have a force between them which is:

[tex]F'=\frac{k(2q_1)(2q_2)}{d^2}=4\frac{kq_1q_2}{d^2}=4F[/tex]

So the new force is 4 times larger than the original force.

Doubling the charge on each particle increases the electric force between them by a factor of 4.

The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.

Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.

So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.

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Answer: Average kinetic energy is greater than gravitational potential energy.

Explanation: The average kinetic energy formulae for a molecule of a gas relative to temperature (in Kelvin) is given below as

E = 3/2 (KT).

Where E = average kinetic energy =?

K = boltzman constant = 1.381×10^-23 m²kg/s²k

T = temperature = 300 k

By substituting the parameters in the formulae, we have that

E = 3/2 ×( 1.381×10^-23 × 300)

E = 1.5 × 4.143×10^-21

E = 6.2145×10^-21 J

To get the gravitational potential energy, we use the fact that

Gravitational potential energy = gravitational energy at the top - gravitational energy at bottom.

At the top, the height of cube is h= 15cm = 0.15m, g = acceleration dude to gravity = 9.8m/s², m = mass of molecule of oxygen = 1.661×10^-27 kg

Gravitational potential energy = mgh = 1.661×10^-27× 9.8 × 0.15 = 2.442×10^-27 J

At the bottom, height is zero, hence gravitational potential energy is also zero.

Hence the final gravitational potential energy = potential energy at top - potential energy at bottom =

2.442×10^-27 - 0 = 2.442×10^-27 J.

Gravitational potential energy = 2.442×10^-27 J

Average kinetic energy = 6.2145×10^-21 J

As we can see that the average kinetic energy is bigger than the gravitational potential energy.

Compare the average kinetic energy of oxygen molecules at 300 K to the change in their gravitational potential energy if they fall from the top to the bottom of a container.

Oxygen (O2) molecules at 300 K have an average kinetic energy of 3kT, where k is Boltzmann's constant. If an oxygen molecule falls from the top of a container to the bottom, its change in gravitational potential energy is mgΔh, where m is the mass of the molecule, g is the acceleration due to gravity, and Δh is the height change.

Answer:

The net force is 300.8 N

Explanation:

∆H = Cp(T1 - T2) = 1/2(V2^2 - V1^2)

Cp is the heat capacity of air at constant pressure = 1005 J/kg

T1 is initial temperature of air = 500 K

T2 is the exit temperature of air = 440 K

V1 is the initial velocity of air = 50 m/s

V2 is the exit velocity of air

1005(500 - 440) = 1/2(V2^2 - 50^2)

60,300×2 = V2^2 - 2,500

V2^2 = 120,600 + 2,500 = 123,100

V2 = sqrt(123,100) = 350.8 m/s

Net force (F) = mass flow rate × change in velocity = 1 kg/s × (350.8 - 50)m/s = 1 kg/s × 300.8 m/s = 300.8 kgm/s^2 = 300.8 N

The net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].

Given:

- Inlet conditions (state 1):

 - [tex]\( P_1 = 12 \) bar = \( 12 \times 10^5 \) Pa[/tex]

 - [tex]\( T_1 = 500 \) K[/tex]

 - [tex]\( v_1 = 50 \) m/s[/tex]

- Exit conditions (state 2):

 - [tex]\( P_2 = 7 \) bar = \( 7 \times 10^5 \) Pa[/tex]

 - [tex]\( T_2 = 440 \) K[/tex]

- Mass flow rate, [tex]\( \dot{m} = 1 \) kg/s[/tex]

1. Calculate specific volumes:

 [tex]\( v_1 = \frac{RT_1}{P_1} = \frac{287 \times 500}{12 \times 10^5} \approx 0.02379 \) m^3/kg[/tex]

  [tex]\( v_2 = \frac{RT_2}{P_2} = \frac{287 \times 440}{7 \times 10^5} \approx 0.02086 \) m^3/kg[/tex]

2. Calculate the change in specific volume:

 [tex]\( \Delta v = v_2 - v_1 = 0.02086 - 0.02379 = -0.00293 \) m^3/kg[/tex]

3. Calculate the net force in the direction of flow:

  [tex]\( F = \dot{m} \cdot \Delta v = 1 \cdot (-0.00293) = -0.00293 \) N[/tex]

Since the force is negative, it indicates that the force is acting opposite to the direction of flow.

Therefore, the net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].

Answer:7.92 N

Explanation:

Given

mass of book [tex]m=1.83\ kg[/tex]

coefficient of static friction [tex]\mu _s=0.442[/tex]

coefficient of kinetic friction [tex]\mu _k=0.240[/tex]

To move the book, one need to overcome  the static friction  

Static friction [tex]F_s=\mu _sN[/tex]

[tex]F_s=\mu _s\times 1.83\times 9.8[/tex]

[tex]F_s=0.442\times 1.83\times 9.8[/tex]

[tex]F_s=7.92\ N[/tex]

After overcoming the Static friction , Force needed to move the block is

[tex]F_k=\mu _kN[/tex]

[tex]F_k=0.240\times 1.83\times 9.8[/tex]

[tex]F_k=4.30\ N[/tex]

Final answer:

The force needed to begin moving a 1.83 kg book on a flat desk, given a coefficient of static friction of 0.442, is approximately 7.90 N. This is calculated using the book's weight and the static friction formula fs(max) = μs * N.

Explanation:

To calculate the force needed to begin moving the book, we need to use the coefficient of static friction and the book's weight. The formula to find the maximum static frictional force (fs max) that must be overcome to start moving the object is fs(max) = μs * N, where μs is the coefficient of static friction and N is the normal force.

The book's weight is given by W = m * g, where m is the mass of the book (1.83 kg) and g is the acceleration due to gravity (9.81 m/s²). The weight of the book is equivalent to the normal force (N) exerted by the desk on the book since the book is resting on a flat surface and there are no other vertical forces acting on it. This simplifies the normal force to N = W = m * g.

Using the given coefficient of static friction (0.442) and the calculated normal force, the force needed to begin moving the book is fs(max) = 0.442 * (1.83 kg * 9.81 m/s²) ≈ 7.90 N. Therefore, a force slightly greater than 7.90 N is required to overcome static friction and start moving the book.

Answer:

Explanation:

The force of friction acting on the system

= .04 x 9.8 ( 10.2 + 7 )

= 6.74 N

Net force = 8.9 - 6.74

= 2.16 N

Acceleration in the system

= 2.16 /  ( 10.2 + 7 )

= .12558 m / s ²

Contact force between boxes = FP

Considering force on box A

Net force = 8.9 - FP

Applying Newton's law on box A

8.9 - FP = 10.2 x .12558

= 1.28

FP = 8.9 - 1.28

= 7.62 N

Answer: elastic (e = 2.43)

Explanation:

The price elasticity formulae is given below as

Elasticy of price = change in quantity demanded / change in price.

P1 =$2.60, P2 = $6.30, q1 =17 and q2 = 8

e = q2 - q1/ P2 - P1

e = 8 - 17/ 6.30 - 2.60

e = - 9 /3.7

e = - 2.43

We take the modulus of e to have a positive value. Hence e = 2.43

Since e is greater than 1, then the elasticity of demand is elastic

Final answer:

The elasticity of demand between the given prices is calculated using the midpoint formula. It results in a value of -0.865 (ignoring the minus sign), indicating that the demand is elastic.

Explanation:

To calculate the elasticity of demand, we can use the midpoint formula, which is defined as the percentage change in quantity demanded divided by the percentage change in price. The formula for computing elasticity is:

Price Elasticity of Demand (Ed) = ([(Q2 - Q1) / ((Q2 + Q1)/2)] / [(P2 - P1) / ((P2 + P1)/2)]) * 100

Using the information provided: At a price of $2.60, quantity demanded is 17, and at a price of $6.30, quantity demanded is 8. We can fill in the values:

(Q2 - Q1) is (8 - 17) = -9

(Q2 + Q1)/2 is (8 + 17)/2 = 12.5

(P2 - P1) is ($6.30 - $2.60) = $3.70

(P2 + P1)/2 is ($6.30 + $2.60)/2 = $4.45

Now putting all these into the formula:

Ed = [(-9 / 12.5) / (3.70 / 4.45)] * 100

The percentage change in quantity is -72%, and the percentage change in price is 83.15%. Therefore:

Ed = (-0.72 / 0.8315) * 100

Ed = -0.865 (ignoring the minus sign)

This value is greater than 1, which indicates that the demand is elastic between these two prices; meaning, the quantity demanded is quite responsive to price changes.

Remember, we ignore the negative sign and use the absolute value when discussing elasticity, even though it is understood that price and quantity demanded move in opposite directions.

Answer:

[tex]\int\limits^2_0 {w} \, dt[/tex]

Explanation:

If Angular velocity w (omega ) is given, which is defined as, w = (change in angle)/(change in unit time).

then simply taking integral from 0 to 2 gives us the answer.

Answer:

Change in theta = 0.793rad

Explanation:

Please see attachment below.

The mass of the North American continent is calculated by multiplying the volume of the continent by the density of the rock it is made of. The volume is determined based on the given dimensions, converted to meters. Multiplying the resulting volume by the rock's density provides the estimated mass of 1.59712e21 kilograms.

In order to find the mass of a continent, we need to multiply the volume of the continent by the density of the rock that it's made of. First, we need to convert the dimensions of the North American continent into meters from kilometers since the density is in kg/m³. That's 4700km x 1000m/km =4.7e6m for length & width, and 25km x 1000m/km = 2.5e4m for depth. Thus, the volume of the continent becomes 4.7e6m x 4.7e6m x 2.5e4m = 5.54e17m³.

Now, we multiply by the rock's density, which is 2880 kg/m³. So, mass = volume * density = 5.54e17 m³ x 2880 kg/m³ = 1.59712e21 kg. So, the theoretical mass of the North American continent is approximately 1.59712e21 kilograms considering its simplified shape as a slab of rock.

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Answer:

a) [tex]-22.5 rad/s^2[/tex]

b) 30.6 revolutions

c) 4.13 s

d) 52.9 m

e) 25.6 m/s

Explanation:

a)

The relationship between linear acceleration and angular acceleration for an object in circular motion is given by

[tex]a=\alpha r[/tex]

where

[tex]a[/tex] is the linear acceleration

[tex]\alpha[/tex] is the angular acceleration

r is the radius of the motion of the object

For the tires of the  car in this problem, we have:

[tex]a=-6.2 m/s^2[/tex] is the linear acceleration (the car is slowing  down, so it is a deceleration, therefore the negative sign)

r = 0.275 m is the radius of the tires

Solving for [tex]\alpha[/tex], we find the angular acceleration:

[tex]\alpha = \frac{a}{r}=\frac{-6.2}{0.275}=-22.5 rad/s^2[/tex]

b)

To solve this part of the problem, we can use the suvat equation for the rotational motion, in particular:

[tex]\omega^2 - \omega_0^2 = 2\alpha \theta[/tex]

where:

[tex]\omega[/tex] is the final angular velocity

[tex]\omega_0[/tex] is the initial angular velocity

[tex]\alpha[/tex] is the angular acceleration

[tex]\theta[/tex] is the angular displacement

Here we have:

[tex]\omega=0[/tex] (the tires come to a stop)

[tex]\omega_0 = 93 rad/s[/tex]

[tex]\alpha = -22.5 rad/s^2[/tex]

Solving for [tex]\theta[/tex], we find the angular displacement:

[tex]\theta=\frac{\omega^2-\omega_0^2}{2\alpha}=\frac{0^2-(93)^2}{2(-22.5)}=192.2 rad[/tex]

And since 1 revolution = [tex]2\pi rad[/tex],

[tex]\theta=\frac{192.2}{2\pi}=30.6 rev[/tex]

c)

To solve this part, we can use another suvat equation:

[tex]\omega=\omega_0 + \alpha t[/tex]

where in this case, we have:

[tex]\omega=0[/tex] is the final angular velocity, since the tires come to a stop

[tex]\omega_0 = 93 rad/s[/tex] is the initial angular velocity

[tex]\alpha=-22.5 rad/s^2[/tex] is the angular acceleration

t is the time

Solving for t, we can find the time required for the tires (and the car) to sopt:

[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{0-93}{-22.5}=4.13 s[/tex]

d)

The car travels with a uniformly accelerated motion, so we can find the distance it covers by using the suvat equations for linear motion:

[tex]s=vt-\frac{1}{2}at^2[/tex]

where:

v = 0 is the final velocity of the car (zero since it comes to a stop)

t = 4.13 s is the time taken for the car to stop

[tex]a=-6.2 m/s^2[/tex] is the deceleration for the car

s is the distance covered during this motion

Therefore, substituting all values and calculating s, we find the distance covered:

[tex]s=0-\frac{1}{2}(-6.2)(4.13)^2=52.9 m[/tex]

e)

The relationship between angular velocity and linear velocity for a rotational motion is given by

[tex]v=\omega r[/tex]

where

v is the linear velocity

[tex]\omega[/tex] is the angular speed

r is the radius of the circular motion

In this problem:

[tex]\omega_0 = 93 rad/s[/tex] is the initial angular speed of the tires

r = 0.275 m is the radius of the tires

Therefore, the initial velocity of the car is:

[tex]u=\omega_0 r = (93)(0.275)=25.6 m/s[/tex] is the initial velocity of the car

Answer:

EMF will induce in the conducting loop when

The magnetic field must be changing

Explanation:

As per Farady's law of EMI we know that the magnetic flux linked with the closed conducting loop must be changed with time

so we have

[tex]EMF = \frac{d\phi}{dt}[/tex]

so we know that

[tex]\phi = BAcos\omega t[/tex]

now if magnetic field is changing with time then we have

[tex]EMF = Acos\theta \frac{dB}{dt}[/tex]

so EMF will induce when magnetic field will change with time

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

[tex]m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}[/tex]

[tex]\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s[/tex]

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Answer:

0.65 A.

Explanation:

Given:

Pmax = 10 W

R = 23.9 Ω

Formula for calculating power,

P = I × V

= I^2 × R

I^2 = 10/23.9

I = 0.65 A.

Using an electromagnet to lift scrap metal is advantageous because it can be turned off to release the metal, and its strength can be adjusted to handle different loads.

Using an electromagnet to move scrap metal is more practical than using a permanent magnet for several reasons. Firstly, an electromagnet can be turned on and off, useful when you want to release the metal after lifting it. This is not possible with a permanent magnet, which would require a physical effort to detach the metal pieces. Secondly, the strength of an electromagnet can be adjusted by controlling the electric current. This allows for strong magnetic effects that can be finely tuned for the weight and type of scrap being lifted. Lastly, industrial electromagnets can be designed to lift thousands of pounds of metallic waste, which might not be feasible with the size and strength of a permanent magnet.

However, there are limits to how strong electromagnets can be made, mainly due to coil resistance leading to overheating. In cases where extremely strong magnetic fields are necessary, such as in particle accelerators, superconducting magnets may be employed, although these also have their limits, since superconducting properties can be destroyed by excessively strong magnetic fields.

Answer: P = 25050.8w

Explanation:

total energy at top = K.E + P.E

= (1/2)(6861)(100) + 6861(9.81)(142)

total energy at bottom

= (1/2)(6861)(48)^2

work done = energy at top - energy at bottom

average velocity = (48+10)/2

time = 2300/average velocity

power = work done/time

plus potential) at the base and the top; is the energy input from the engine

the ascent time is the average speed, (top + bottom) / 2; divided by the 2.3 km distance

energy / time equals power

Answer:

a) [tex]||\vec F_{B}||=62.664\,N[/tex], b) From east to west.

Explanation:

Vectorially, the magnetic force can be calculed by the following formula:

[tex]\vec F_{B} = i\cdot \vec l\, \times \, \vec B[/tex]

The cross product is:

[tex]\vec F_{B} = \left|\begin{array}{ccc}i&j&k\\1015000\,A\cdot m&0\,A\cdot m&0\,A\cdot m\\22.471\times 10^{-6}\,T&-61.738\times 10^{-6}\,T&0\,T\end{array}\right|[/tex]

[tex]\vec F_{B} = - 62.664\,N\cdot k[/tex]

a) The magnitude of the magnetic force is:

[tex]||\vec F_{B}||=62.664\,N[/tex]

b) The direction of the magnetic force is:

From east to west.

Answer:

[tex]\phi=6.28\times10^{-3}\;\;weber[/tex]

Explanation:

Given,

Magnetic field [tex]B=0.2\;\;T\\[/tex]

Radius [tex]r=0.1\;\;m[/tex]

The angle between the area vector and magnetic field is 0 degree, because the direction of area vector is always perpendicular to the plane.

[tex]\phi=BAcos\theta\\\phi=o.2\times\pi (0.1)^2\times cos0^o\\\phi=0.00628\;\;weber\\\phi=6.28\times10^{-3}\;\;weber[/tex]

The magnetic flux through a flat circular loop rotating in a uniform magnetic field is zero when the plane of the loop and the magnetic field vector are perpendicular.

To find the magnetic flux through the loop when the plane of the loop and the magnetic field are perpendicular, we will need to use the formula for magnetic flux:

Φm = BA cos θ

where Φm is the magnetic flux through the surface, B is the magnitude of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field direction and the normal (perpendicular) to the surface. In this case, if the loop and the magnetic field are perpendicular, θ = 90°, and cos 90° = 0. Hence, the magnetic flux will be zero.

However, if the provided problem included a different angle, we would adjust the equation to accommodate that. For example, if the field were parallel to the loop (θ = 0), then cos θ would equal 1 and the magnetic flux would just be the product of B and A (A = πr²)

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Answer: the mutual inductance is - 0.0978H

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

M = 0.1H

Explanation:

Please see attachment below.

Answer:

Explanation:

As there is no external torque is applied so the angular momentum remains constant.

L = I ω = constant

Where, I is the moment of inertia of the system and ω is the angular velocity

As we move towards the edge, the moment of inertia increases, hence the angular velocity decreases.

Answer:

500 m/s

Explanation:

Momentum, p is a product of mass and velocity. From law of conservation of momentum, the initial momentum equals final momentum

[tex]m_1v_1=m_2v_2\\1000*10=20v_2\\v_2=10000/20=500 m/s[/tex]

Here m and v represent mass and velocity respectively and subscripts 1 and 2 represent car and barrier respectively

Therefore, the velocity of barrier will be 500 m/s