Which of the following is a TRUE statement?
a. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one to a hot one, depending on whether or not the process is reversible or irreversible.
b. It is not possible to convert work entirely into heat.
c. The second law of thermodynamics is a consequence of the first law of thermodynamics.
d. It is impossible to transfer heat from a cooler to a hotter body.
e. All of these statements are false.

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

The average velocity of the sled, given by the formula v_avg = s / t, evaluates the distance covered over a certain duration of time. The sled has moved a distance 's' during a time interval 't', and so its average velocity over this interval is simply 's' divided by 't'. The presence of friction does not affect the calculation of average velocity.

The average velocity of the sled is simply the distance traveled over a certain time interval. In this case, the sled starts from rest and is being pulled by a horizontal force F against a frictional force. Average Velocity, v_avg, is given by the formula:

v_avg = s / t

In physics, velocity is a measure of the rate of change of position concerning time, so the average velocity is simply the total distance (or displacement) divided by the total time. Given the fact that the sled has moved a distance, s, during the time interval t, the average velocity is simply s divided by t.

With this definition, the sled's average velocity can be determined without needing to tabulate its speed or direction throughout the time interval even in the presence of friction.

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The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g.

The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g, where g is the acceleration due to gravity.

The mass of the book does not matter in this scenario. The coefficient of static friction is a property that depends on the nature of the surfaces in contact, and it determines the maximum force of static friction that can act between the book and the wagon before the book starts to slide down. The acceleration required to prevent sliding is independent of the mass of the book.

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Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

[tex]\sigma_{max}[/tex] = Maximum stress

[tex]\sigma_{min}[/tex] = Minimum stress

[tex]\sigma_m[/tex] = Mean stress = 50 MPa

[tex]\sigma_a[/tex] = Stress amplitude = 225 MPa

Mean stress is given by

[tex]\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}[/tex]

Stress amplitude is given by

[tex]\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa[/tex]

[tex]\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa[/tex]

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

[tex]R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636[/tex]

The stress ratio is -0.63636

Stress range is given by

[tex]\sigma_{max}-\sigma_{min}=450\ MPa[/tex]

Magnitude of the stress range is 450 MPa

The maximum and minimum stress levels could be calculated as 275 MPa and -175 MPa respectively. The stress ratio comes out to be -0.64, and the magnitude of the stress range is 450 MPa.

In material fatigue studies, the stress levels in a material are often evaluated. In your case:

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Answer:

The loop penetrate 4 cm into the magnetic field.

Explanation:

Given that,

Width w= 5 cm

Length L= 10 cm

mass m = 40 g

Resistance R = 20 mΩ

Initial velocity = 1 m/s

Magnetic field = 2 T

We need to calculate the induced emf

Using formula of emf

[tex]\epsilon=v_{0}Bw[/tex]

Put the value into the formula

[tex]\epsilon =1\times2\times5\times10^{-2}[/tex]

[tex]\epsilon =10\times10^{-2}\ volt[/tex]

We need to calculate the current

Using Lenz's formula

[tex]i=\dfrac{\epsilon}{R}[/tex]

[tex]i=\dfrac{10\times10^{-2}}{20\times10^{-3}}[/tex]

[tex]i=5\ A[/tex]

We need to calculate the force

Using formula of force

[tex]F=i(\vec{w}\times\vec{B})[/tex]

[tex]F=iwB[/tex]

Put the value into the formula

[tex]F=5\times5\times10^{-2}\times2[/tex]

[tex]F=0.5\ N[/tex]

We need to calculate the acceleration

Using formula of acceleration

[tex]a=\dfrac{F}{m}[/tex]

Put the value in to the formula

[tex]a=\dfrac{0.5}{40\times10^{-3}}[/tex]

[tex]a=12.5\ m/s^2[/tex]

We need to calculate the distance

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0-1^2}{2\times(-12.5)}[/tex]

[tex]s=0.04\ m[/tex]

[tex]s=4\ cm[/tex]

Hence, The loop penetrate 4 cm into the magnetic field.

To develop this problem it is necessary to apply the concepts developed by Clausius - Claperyron.

This duet found the relationship between temperature and pressure expressed as,

[tex]ln P = constant - \frac{\Delta H}{RT}[/tex]

For the two states that we have then we could define the pressure and temperature in each of them as

[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]

Where,

[tex]P_{1,2}[/tex]= Pressure at state 1 and 2

[tex]T_{1,2}[/tex]= Temperature at state 1 and 2

[tex]\Delta H[/tex]= Enthalpy of Vaporization of a substance

R = Gas constant (8.134J/mol.K)

Our values are given by,

[tex]P_1 = 1atm \\\Delta H = 38.56*10^{-3} J/mol \\R = 8.134J/mol.K\\T_1 = 78.4\°C = 351.55K\\T_2 =38.8\°C = 311.95K[/tex]

Therefore replacing we have that,

[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]

[tex]ln(\frac{P_2}{1atm}) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]

[tex]ln(P_2) - Ln(1atm) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]

[tex]P_2 = e^{\frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})}[/tex]

[tex]P_2 = 0.187355atm[/tex]

Therefore the pressure of the Ethanol at 38.8°C is 0.187355atm

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

b = Wien's displacement constant = [tex]2.897\times 10^{-3}\ mK[/tex]

T = Temperature

From Wien's law we have

[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{2900}\\\Rightarrow \lambda_m=9.98966\times 10^{-7}\ m[/tex]

Frequency is given by

[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{9.98966\times 10^{-7}}\\\Rightarrow \nu=3.00311\times 10^{14}\ Hz[/tex]

For Halogen

[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{3400}\\\Rightarrow \lambda_m=8.52059\times 10^{-7}\ m[/tex]

Frequency is given by

[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{8.52059\times 10^{-7}}\\\Rightarrow \nu=3.52088\times 10^{14}\ Hz[/tex]

The maximum frequency is produced by Halogen bulbs which is closest to the value of [tex]5.5\times 10^{14}\ Hz[/tex]

Ratio

[tex]\frac{3.00311\times 10^{14}}{3.52088\times 10^{14}}=0.85294[/tex]

The ratio of Incandescent to halogen peak frequency is 0.85294

To solve this problem it is necessary to apply the concepts related to electromotive force or induced voltage.

By definition we know that the induced emf in the loop is equal to the negative of the change in the magnetic field, that is,

[tex]\epsilon = -A \times \frac{\Delta B}{\Delta t}[/tex]

[tex]\epsilon = -A \times (\frac{B_f-B_i}{t_f-t_i})[/tex]

Where A is the area of the loop, B the magnetic field and t the time.

Replacing with our values we have that

[tex]\epsilon = -(\pi (1.5*10^{-2})^2)(\frac{0-23*10^{-6}}{7*10^{-3}-0})[/tex]

[tex]\epsilon = 2.3225*10^{-6}V[/tex]

Therefore the thermal energy produced is given by

[tex]E = P*t = \frac{\epsilon^2}{R}t[/tex]

[tex]E = \frac{(2.3225*10^{-6})^2}{8*10^{-6}}*(7*10^{-3})[/tex]

[tex]E = 4.719*10^{-9}J[/tex]

The thermal energy produced in the loop is [tex]4.719*10^{-9}J[/tex]

Answer:

The activity is 811.77 Bq

Solution:

As per the question:

Half life of Thorium, [tex]t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs[/tex]

Mass of Thorium, m = 200 mg = 0.2 g

M = 232 g/mol

Now,

No. of nuclei of Thorium in 200 mg of Thorium:

[tex]N = \frac{N_{o}m}{M}[/tex]

where

[tex]N_{o }[/tex] = Avagadro's number

Thus

[tex]N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}[/tex]

Also,

Activity is given by:

[tex]\frac{0.693}{t_{\frac{1}{2}}}\times N[/tex]

[tex]A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr[/tex]

[tex]A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq[/tex]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   [tex]V = \epsilon - Ir[/tex]

now,

   [tex]IR = \epsilon - Ir[/tex]

   [tex]I(R+r) = \epsilon[/tex]

   [tex]I= \dfrac{\epsilon}{R+r}[/tex]

inserting the values

   [tex]I= \dfrac{12}{0.05+0.01}[/tex]

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 [tex]I= \dfrac{\epsilon}{R+r}[/tex]

   [tex]I= \dfrac{12}{0.14+0.01}[/tex]

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W

Answer:

Explanation:

Resistance of motor, R = 0.05 ohm

internal resistance of battery, r = 0.01 ohm

Voltage of battery, V = 12 V

(a) Total resistance, R' = R + r = 0.05 + 0.01 = 0.06 ohm

Let the current be i.

use Ohm's law

i = V / R'

i = 12 / 0.06 = 200 A

(b) Voltage across motor, V' = i x R = 200 x 0.05 = 10 V

(c) Power, P = i²R = 200 x 200 x 0.05 = 2000 Watt.

(d) Total resistance, R' = 0.05 + 0.1 + 0.09 = 0.15 ohm

i = V / R' = 12 / 0.15 = 80 A

V' = i x R = 80 x 0.05 = 4 V

P' = i²R = 80 x 80 x 0.05 = 320 Watt

Answer:

a.) 5.24 10⁻³ N . b) 0.013 N

Explanation:

a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.

We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.

In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.

The only force that has horizontal and vertical components, is the tension in the string.

We can apply Newton's 2nd Law to both directions, as follows:

T cos 24º - mg = 0

-T sin 24º + Fe = 0

where T= Tension in the string.

Fe = Electrical Force

mg = Fg = gravity force

⇒ T = mg/ cos 24º

Replacing in the horizontal forces equation:

-mg/cos 24º . sin 24º = -Fe

∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N

b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:

T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N

Final answer:

To find the magnitude of the electrical force (F⃗ elec) exerted on the plastic ball, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The electrical force is the only horizontal force acting on the ball, so it must be balanced by the horizontal component of the tension in the string.

Explanation:

The horizontal component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the horizontal component of the tension is:

T_horizontal = T * cos(24.0 degrees)

Since the ball is in equilibrium, the magnitude of the electrical force is equal to the horizontal component of the tension:

|F⃗ elec| = T_horizontal

To find the tension in the string, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The gravitational force (F⃗ g) acting on the ball is balanced by the vertical component of the tension in the string.

The vertical component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the vertical component of the tension is:

T_vertical = T * sin(24.0 degrees)

Since the ball is in equilibrium, the magnitude of the gravitational force is equal to the vertical component of the tension:

|F⃗ g| = T_vertical

The gravitational force can be calculated using the mass of the ball (m) and the acceleration due to gravity (g):

|F⃗ g| = m * g

Substituting the expression for the vertical component of the tension into the equation for the gravitational force, we get:

m * g = T * sin(24.0 degrees)

Solving for the tension (T), we get:

T = (m * g) / sin(24.0 degrees)

Substituting the given values for the mass of the ball (m) and the acceleration due to gravity (g), we get:

T = (1.20 g * 9.8 m/s^2) / sin(24.0 degrees)

T ≈ 5.10 N

b. The tension in the string is approximately 5.10 N.

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

The mass of the load is 2636.8 kg

Mass of the truck m = 7100 kg

Angle θ = 15°

velocity v = 15m/s

Acceleration a = 1.5 m/s²

Now let us assume that the mass of the truck is m₁ and the mass of the load is m₂. The thrust from the engine be T.

Initially, the thrust (T) balances the total weight:

[tex]T = (m_1+m_2) g sin\theta[/tex]

When the load falls off, the thrust (T) remains the same

so the net force on the truck

[tex]F = T -m_1gsin\theta[/tex]

[tex]T-m_1gsin\theta = m_1a \\\\ T = m_1(a + gsin\theta)[/tex]

comparing both the equation for T we get:

[tex](m_1+m_2)gsin\theta = m_1(a + gsin\theta)\\\\m_1gsin\theta + m_2gsin\theta =m_1a + m_1gsin\theta\\\\m_2gsin\theta = m_1a[/tex]

Now, m₁+m₂ = 7100kg

m₁= 7100 – m₂

[tex]m_2gsin\theta = (7100-m_2)a\\\\m_2gsin\theta = 7100a-m_2a\\\\m_2gsin\theta + m_2a = 7100a\\\\m_2 (gsin\theta + a) = 7100a\\\\m_2 = 7100a/(gsin\theta + a)\\\\m_2 = (7100 \times1.5) / (9.8sin(15) + 1.5)\\\\m_2 = 2636.8 kg[/tex]

The load has a mass of 2636.8 kg

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Answer

given,                                    

radius of tire = 25 cm = 0.25 m

initial velocity = 0 m/s                

final velocity = 30 m/s                  

time = 10 s                                        

a) acceleration with respect to center  

   v = u + a t                      

   30 = 0 + 10 a                      

    a = 3 m/s²                            

   radial acceleration            

  [tex]a_r = \dfrac{v^2}{r}[/tex]      

  [tex]a_r = \dfrac{2^2}{0.25}[/tex]

  [tex]a_r =16\ m/s^2[/tex]

  [tex]a=\sqrt{a_r^2 + a_t^2}[/tex]

  [tex]a=\sqrt{16^2 + 3^2}[/tex]

  [tex]a=16.28\ m/s[/tex]

acceleration w.r.t center of tire = [tex]a=16.28\ m/s[/tex]

b) acceleration with respect to road will be equal to 3 m/s²

Answer:

L = 0.0319 H

Explanation:

Given that,

Number of loops in the solenoid, N = 900

Radius of the wire, r = 3 cm = 0.03 m

Length of the rod, l = 9 cm = 0.09 m

To find,

Self inductance in the solenoid

Solution,

The expression for the self inductance of the solenoid is given by :

[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]

[tex]L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}[/tex]

L = 0.0319 H

So, the self inductance of the solenoid is 0.0319 henries.

Since there is a positive work on an object, this directly indicates that there is an increase in Energy on the body.

Work can be expressed in terms of Force and distance as

W = Fd

Where,

F = Force

d = Distance

The force in turn is a description of how mass accelerates according to Newton's second law, that is

F = ma

The mass (m) remains constant, but the acceleration (a) is constant, which implies directly that there is a gradual change in the velocity of the object.

Therefore the correct answer is: The object is speeding up.

The work is positive so the energy of the object is  increasing so the object is speeding up

As we know that the work is the product of the force and the displacement

[tex]W=F\times D[/tex]

Where,

F = Force

D= Distance

And from newtons second law we can see that

[tex]F=m\times a[/tex]

Since here mass will be constant to there will be a change in the velocity that is acceleration in the body so the energy of the body will change

Thus work is positive so the energy of the object is  increasing so the object is speeding up

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Answer:

h=12.04m

Explanation:

1) Notation and some important concepts

[tex]\omega_i[/tex] represent the initial angular velocity

[tex]v_i[/tex] represent the initial velocity

[tex]h[/tex] represent the final height reached by the mass

[tex]M[/tex] represent the mass of the object

[tex]W_f[/tex] represent the work due the friction force (variable of interest)

[tex]KE_{rot}[/tex] represent the rotational energy

[tex]KE_{tra}[/tex] represent the transitional kinetic energy

[tex]PE=mgh[/tex] represent the potential energy

[tex]I=0.8MR^2[/tex] represent the rotational inertia

W= 395 N is the weight of the object

For this problem we can use the principle of energy conservation, this principle states that "the total energy of an isolated system remains constant; it is said to be conserved over time".

At the begin the wheel had rotational energy defined as "The kinetic energy due to rotational motion. Is a scalar quantity and have units of energy usually Joules". And this energy is represented by the following formula: [tex]KE_{rot}=\frac{1}{2}I\omega^2_i[/tex]

At the starting point the wheel also had kinetic energy defined as "The energy of mass in motion" and is given by the formula : [tex]KE_{tran}=\frac{1}{2}mv_i^2_i[/tex]

At the end of the movement we have potential energy since the mass is at height h the potential energy is defined as "The energy stored within an object, due to the object's position, arrangement or state" and is given by the formula [tex]PE=mgh[/tex].

Since we have friction acting we have a work related to the force of friction and we need to subtract this from the formula of conservation of energy

2) Formulas to use

The figure attached is an schematic draw for the problem

Using the principle of energy conservation we have:

[tex]KE_{rot}+KE_{tran}-W_f =PE[/tex]

Replacing the formulas for each energy w ehave:

[tex]\frac{1}{2}I\omega^2_i+\frac{1}{2}mv_i^2_i-W_f =Mgh[/tex]    (1)

We also know that [tex]v_i =\omega_i R[/tex] and [tex]I=0.8MR^2[/tex] so if we replace this into equation (1) we got:

[tex]0.8(\frac{1}{2})M(R\omega_i)^2 +\frac{1}{2}M(\omega_i R)^2-W_f=Mgh[/tex]    (2)

We also know that the weight is defined as [tex]W=mg[/tex] so then [tex]M=\frac{W}{g}[/tex], so if we replace this into equation (2) we have:

[tex]0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f=Wh[/tex]    (3)

So then if we solve for h we got:

[tex]h=\frac{0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f}{W}[/tex]    (4)

3) Solution for the problem

Now we can replace the values given into equation (4):

[tex]h=\frac{0.8(\frac{1}{2})\frac{395N}{9.8\frac{m}{s^2}}(0.652m(23.1\frac{rad}{s}))^2 +\frac{1}{2}\frac{395}{9.8\frac{m}{s^2}}(23.1\frac{rad}{s}(0.652m))^2-3470J}{395N}=12.04m[/tex]    (4)

So then our final answer would be h=12.04m

Answer:

0.25938 W

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]\nu[/tex] = Frequency = 3 GHz

d = Diameter of lossless antenna = 1 m

r = Radius = [tex]\frac{d}{2}=\frac{1}{2}=0.5\ m[/tex]

[tex]A_t[/tex] = Area of transmitter

[tex]A_r[/tex] = Area of receiver

R = Distance between the antennae = 40 km

[tex]P_r[/tex] = Power of receiver = [tex]10\times 10^{-9}\ W[/tex]

[tex]P_t[/tex] = Power of Transmitter

Wavelength

[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{3\times 10^9}\\\Rightarrow \lambda=0.1\ m[/tex]

From Friis transmission formula we have

[tex]\frac{P_t}{P_r}=\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=P_r\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=10\times 10^{-9}\frac{0.1^2\times (40\times 10^3)^2}{\pi 0.5^2\times \pi 0.5^2}\\\Rightarrow P_t=0.25938\ W[/tex]

The power that should be transmitted is 0.25938 W

Answer: It is not gold. The density is 1.25 g/cm3, very little compared with gold.

Explanation:

According to Archimedes’ principle, any body submerged in a liquid, receives an upward force, called buoyant force, which magnitude is equal to the weight of the liquid that displaces.

This weight can be calcultated as follows:

Fb= δH2O . Vbody g

We are told that the volume of the liquid displaced (equal to the volume of the body as it is completely submerged in water) is 400 cm3.

If we know that the mass of the bracelet is 0.5 Kg, we can find out the density of the material, just applying the same relationship between mass, density and volume:

m = δbracelet. V  

δbracelet = m / V = 500 g / 400 cm3= 1.25 g/cm3

As the gold density is approximately 19.3 g/cm3, it is clear that the bracelet is not made from gold, but from something less dense.

The density of Vadim's bracelet is calculated to be 1.25 g/cm³, which is significantly lower than the density of pure gold (19.3 g/cm³). Thus, the bracelet is not made of gold and could be made from a lighter metal or alloy.

To determine if Vadim's bracelet is gold, we must calculate its density and compare it to the density of pure gold. The density of a substance is the mass-to-volume ratio, and for gold, it is 19.3 g/cm³. Given that the bracelet has a mass of 0.5 kg (or 500 g) and a volume displacement of 400 cm³ when submerged in water, we can calculate its density using the formula:

Density = Mass / Volume

density = 500 g / 400 cm³ = 1.25 g/cm³

The calculated density of 1.25 g/cm³ is significantly lower than the density of pure gold, which suggests that the bracelet is not made of gold. Since the density is lower than that of gold, the bracelet could be made of a metal or alloy with a lower density, such as aluminium or a mixture that includes materials with lighter densities.

The television set moves sideways approximately 0.15 µm.

To calculate the distance the television set moves sideways, we need to use the equation for shear deformation. The equation is Ax = (F * L0 * L2) / (S * A), where Ax is the displacement, F is the force applied, L0 is the original length, L2 is the height of the rubber pads, S is the shear modulus, and A is the cross-sectional area. Substituting the known values, we have Ax = (200 N * 1.0 cm) / (2.6 x 10^6 N/m^2 * (3.14 x (0.0060 m)^2)). Solving for Ax, we find Ax ≈ 1.50 x 10^-7 m, or 0.15 µm.

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Answer:

Explanation:

Kinetic energy of ball

= .5 x .5 x 20²

= 100 J

Original kinetic energy of door = 0

Total kinetic energy before ball hitting the door

= 100 J

We shall apply law of conservation of momentum to calculate angular velocity of the door after ball hitting it.

change in angular  momentum of ball

= mvr - mur , u is initial velocity and v is final velocity of ball

= .5 ( 20 + 16 ) x .06

= 1.08

Change in angular momentum of door

= I x ω - 0

1/3 x 35 x .09² x ω

= .0945 x ω

so

.0945 x ω = 1.08

ω = 11.43

rotational K E of door after collision

= 1/2 I ω²

= .5 x .0945 x 11.43 ²

= 6.17 J  

Kinetic energy of ball after collision

= 1/2 x .5 x 16²

= 64

Total KE of door and ball

= 64 + 6.17

= 70.17 J

LOSS OF ENERGY

= 100 - 70.17 J

= 29.83 J

Answer:

a) V = 465.9 m/s

b) θ = 70.529°

Explanation:

Let's first calculate angular velocity of earth:

[tex]\omega=\frac{2\pi}{23.9h}*1h/3600s[/tex]

Velocity of a person on Ecuador will be:

[tex]V_E = \omega*R[/tex]

[tex]V_E = 465.9 m/s[/tex]

For part b, since angular velocity is the same:

[tex]\frac{\omega*R}{3}=\omega*(R*cos\theta )[/tex]

Solving for θ:

[tex]\theta=acos(1/3)[/tex]

[tex]\theta=70.529\°[/tex]

(a)   speed at the equator: 465.9 m/s

(b) Latitude for 1/3 speed: 50.6°

(a) The tangential speed of a person living in Ecuador is equal to the circumference of the Earth at the equator divided by the period of rotation. The circumference of the Earth at the equator is 2πr, where r is the radius of the Earth. The period of rotation is 23.9 hours, which is equal to 23.9 × 3600 seconds. Therefore, the tangential speed of a person living in Ecuador is:

v = 2πr / T = 2π(6.38 ×[tex]10^6[/tex] m) / (23.9 × 3600 s) ≈ 465.9 m/s

(b) The tangential speed of a person living at a latitude of θ is equal to vr * cos(θ), where vr is the tangential speed of a person living in Ecuador and r is the radius of the Earth. Therefore, we can solve for θ:

cos(θ) = vr / v = 465.9 m/s / v

Solving for θ, we get:

θ = arccos(vr / v) ≈ 50.6°

Therefore, the latitude at which the tangential speed is one-third that of a person living in Ecuador is approximately 50.6°.

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Answer:

[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]

Explanation:

Using the equation to find the fraction does the radius of the sphere change when it is exposed to Venusian atmosphere.

[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{V}{V_0}[/tex]

Replacing to find the relation between the radius

[tex]\frac{V}{V_0}=-\frac{P}{B}[/tex]

Replacing numeric to find the relation

[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{8.9x10^6pa}{6.7x10^{10}pa}[/tex]

[tex]\frac{r}{r_0}=\frac{1}{3}*-1.33x10^{-4}[/tex]

So the relation is

[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]

Answer:

Angular displacement will be [tex]\frac{1}{4}\Theta[/tex]

So option (b) will be the correct option

Explanation:

We have given that firstly object is at rest

So [tex]\omega _i=0rad/sec[/tex]

From law of motion we know that angular displacement is given by

[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2[/tex]

Now angular displacement by the object in [tex]\frac{t}{2}sec[/tex]

[tex]\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta[/tex]

So option (b) will be the correct option

The angle the object rotate through in the time [tex]\frac{1}{2} t[/tex] is [tex]\frac{1}{4} (\theta)[/tex]

Given the following data:

To determine the angle the object rotate through in the time [tex]\frac{1}{2} t[/tex]:

Mathematically, angular displacement is given by this formula:

[tex]\theta = \omega_i t +\frac{1}{2} \alpha t^2[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\theta = 0( t )+\frac{1}{2} \alpha t^2\\\\\theta = \frac{1}{2} \alpha t^2[/tex]

when t = [tex]\frac{1}{2} t[/tex]:

[tex]\theta = \frac{1}{2} \alpha (\frac{t}{2} )^2\\\\\theta = \frac{1}{2} \alpha (\frac{t^2}{4} )\\\\\theta =\frac{1}{4} (\frac{1}{2} \alpha t^2)\\\\\theta =\frac{1}{4} (\theta)[/tex]

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Answer:

Frequency, f = 3.35 Hz

Explanation:

It is given that,

Mass of the object, m = 8 kg

Stretching in the spring, x = 2.2 cm

When the spring is hanged up in the Earth's gravitational field, its weight is balanced by the force in the spring. So,

[tex]mg=kx[/tex]

k is the spring constant

[tex]k=\dfrac{mg}{x}[/tex]

[tex]k=\dfrac{8\times 9.8}{2.2\times 10^{-2}}[/tex]            

k = 3563.63 N/m

Let f is the frequency of oscillation. Its expression is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{3563.63}{8}}[/tex]        

f = 3.35 Hz

So, the frequency of oscillation by the spring is 3.35 Hz. Hence, this is the required solution.                                                                                                                                                                                                            

Answer:

The radius of the second coil will be 7.5 cm

Explanation:

We have given that radius of the first coil [tex]r_1=5.1cm=0.051m[/tex]

Peak emf of the first generator [tex]e_1=1.8Volt[/tex]

Peak emf of the second generator [tex]e_2=3.9Volt[/tex]

We know that induced emf is rate of change of magnetic flux , that is equal to magnetic field multiply by change in area

So emf will be proportional to the change in area

So [tex]\frac{e_1}{e_2}=\frac{A_1}{A_2}[/tex]

[tex]\frac{1.8}{3.9}=\frac{0.051^2}{r_2^2}[/tex]

[tex]r_2=0.075m = 7.5cm[/tex]

So the radius of the second coil will be 7.5 cm

Answer:

7.85 m/s^2

Explanation:

linear or tangential acceleration= dv/dt

⇒[tex]a_t= \frac{12.5-10}{3}[/tex]

=0.83 m/s^2

radial acceleration is given by = [tex]\frac{v^2}{r}[/tex]

⇒[tex]a_r =\frac{12.5^2}{20}[/tex]

= 7.81 m/s^2

total acceleration

[tex]a_T= \sqrt{a_t^2+a_r^2}[/tex]

putting values we get

[tex]a_T= \sqrt{0.83^2+7.81^2}[/tex]

= 7.85 m/s^2

Answer:

Explanation:

Tangential acceleration = ( 12.5 - 10 )/ 3

a_t= .833 m /s²

radial acceleration

= v² / R

12.5² / 20 ( 12.5 m/s is the velocity when it leaves the curve )

a_r= 7.81 ms⁻²  

Total acceleration √( .833² + 7.81²)

=  √( 61.6939)

= 7.85 m/s⁻

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

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Answer:

Explanation:

Given

inclination [tex]=\beta [/tex]

Assuming radius of sphere is r

Now from Free Body Diagram

[tex]Mg\sin \theta -f_r=Ma[/tex]

where [tex]f_r=friction\ force[/tex]

[tex]a=acceleration\ of\ system[/tex]

Now friction will Provide the Torque

[tex]f_r\times r=I\cdot \alpha [/tex]

where [tex]I=moment\ of\ inertia[/tex]

[tex]\alpha =angular\ acceleration [/tex]

[tex]f_r\times r=\frac{2}{3}Mr^2\times \frac{a}{r}[/tex]

in pure rolling [tex]a=\alpha r[/tex]

[tex]f_r=\frac{2}{3}Ma[/tex]

[tex]mg\sin \beta -\frac{2}{3}Ma=Ma[/tex]

[tex]Mg\sin \beta =\frac{5}{3}Ma[/tex]

[tex]a=\frac{3g\sin \beta }{5}[/tex]

Answer:

First bright fringe is at 1.82 mm

First dark fringe is at 2.83 mm

Solution:

As per the question:

Slit width, d = 1.19 mm = [tex]1.19\times 10^{- 3}\ m[/tex]

Distance from the screen, x = 3.53 m

Wavelength of the light, [tex]\lambda = 635\ nm = 635\times 10^{- 9}\ m[/tex]

Now,

We know that the 1st bright fringe from the central fringe is given by:

[tex]y = \frac{n\lambda x}{d}[/tex]

where

n = 1

[tex]y = \frac{1\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 1.88\ mm[/tex]

Also, we know that the 1st dark fringe from the central fringe is given by:

[tex]y = \frac{(n + \frac{1}{2})\lambda x}{d}[/tex]

[tex]y = \frac{(1 + \frac{1}{2})\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 2.83\ mm[/tex]

Answer:

3 cm

Explanation:

To answer this question the equation of continuity can be used

For us to use the equaition of continuity, we will make a few assumptions:

[tex]A_{1}v_{1} = A_{2} v_{2} \\\pi (3cm)^{2} * 5m/s = \pi r^{2} *20m/s\\\\r = 1.5 cm\\D = 2r = 2*1.5cm=3 cm[/tex]

Answer:

Explanation:

mass, m = 86 g

Amplitude, A = 1.3 mm

Maximum acceleration, a = 6.8 x 10^3 m/s^2

(a) Let T be the period of motion and ω be the angular frequency.

maximum acceleration, a = ω²A

6.8 x 10^3 = ω² x 1.3 x 10^-3

ω = 2287.0875 rad/s

T = 2π/ω

T = ( 2 x 3.14) / 2287.0875

T = 2.75 x 10^-3 second

T = 2.75 milli second

(b) maximum speed = ωA

v = 2287.0875 x 1.3 x 10^-3

v = 2.97 m/s

(c) Total mechanical energy

T = 1/2 mω²A²

T = 0.5 x 86 x 10^-3 x 2287.0875 x 2287.0875 x 1.3 x 1.3 x 10^-6

T = 0.38 J

(d) At maximum displacement, the acceleration is maximum

F = m x a = 86 x 10^-3 x 6.8 x 1000

F = 541.8 N

(e) acceleration a = ω² y = ω² x A / 2

a =  2287.0875 x 2287.0875 x 1.3 x 10^-3 / 2

a = 3400 m/s^2

Force, F = 86 x 10^-3 x 3400

F = 292.4 N