The force of attraction between a star of mass M and a planet of mass m (where т «М) is: ЗЫ? a F where is the angular momentum of the planet and a,b are both positive constants. [Note: black hole, in the non- this does approximate the force of attraction between a planet and a relativistic limit, with a = GMm. (a) 15 points Under what conditions is a stable circular orbit possible? Give the radius of the stable circular orbit in terms of the given parameters (M, m,a, b, l). function of a (b) 15 pointsWhat is the smallest radius possible for any circular orbit as a and b, allowing for arbitrary l? (Hint: this occurs in the limit of very large .) Is this circular orbit stable or unstable? (c) [10 points] If the planet travels in a slightly non-circular orbit about a stable radius, find expression for the angular frequency of small radial oscillations

Answer:

No force acts on it. It will not be pushed.

Explanation:

The proton moving in the -X direction implies its velocity is in the -X direction and the magnetic field is given to be in the +X direction. So the angle between them is 180 degrees. So no force acts on the proton, since the velocity and magnetic field are anti parallel.

Magnetic force is given by the equation F = q v B sin theta where theta is the angle between the velocity and the magnetic field. This force will be a maximum if they are perpendicular to each other.

When a proton moves in a uniform magnetic field, the resulting force is perpendicular to the proton's velocity and the magnetic field, pushing the proton in the +y direction.

When a proton moves in the -x direction through a uniform magnetic field in the +x direction, the resulting magnetic force on the proton will be perpendicular to both the velocity of the proton and the magnetic field. This means the proton will be pushed in the +y direction.

Answer:

Option A is the correct answer.

Explanation:

When the frequency of the driving force equals the natural frequency of the system, the system is said to be in resonance. At resonance the system vibrates in maximum amplitude.

Marching soldiers are cautioned to break stride on a bridge is because of resonance, if the frequency of soldiers stride is equal to frequency of bridge, the bridge will vibrate with maximum amplitude. This will in turn collapse bridge.

Option A is the correct answer.

Answer:

0.587 m/s

Explanation:

m = mass of the object = 2.7 kg

k = spring constant = 280 N/m

[tex]w[/tex] = angular frequency

Angular frequency is given as

[tex]w =\sqrt{ \frac{k}{m}}[/tex]

[tex]w =\sqrt{ \frac{280}{2.7}}[/tex]

[tex]w[/tex] = 10.2 rad/s

x = position relation to equilibrium position = 0.020 m

A = amplitude

[tex]v[/tex]  = speed at position "x" = 0.55 m/s

speed is given as

[tex]v = w\sqrt{A^{2} - x^{2}}[/tex]

[tex]0.55 = (10.2)\sqrt{A^{2} - 0.02^{2}}[/tex]

A = 0.0575 m

[tex]v_{max}[/tex] = maximum speed of the object

maximum speed of the object is given as

[tex]v_{max}=A w[/tex]

[tex]v_{max}=(0.0575) (10.2)[/tex]

[tex]v_{max}[/tex] = 0.587 m/s

The maximum speed attained by the object,

[tex]\rm v_m_a_x=0.587\;m/sec[/tex]

Given :

Mass of the object, m = 2.7 Kg

Spring constant, K = 280 N/m

Speed = 0.55 m/sec

Solution :

We know that the angular velocity is given by,

[tex]\rm \omega = \sqrt{\dfrac{K}{m}}[/tex]

[tex]\rm \omega = \sqrt{\dfrac{280}{2.7}}[/tex]

[tex]\rm \omega = 10.2\;rad/sec[/tex]

Now, speed is given as

[tex]\rm v = \omega\sqrt{A^2-x^2}[/tex]

[tex]0.55=(10.2)\sqrt{A^2-0.02^2}[/tex]

[tex]\rm A = 0.0575\;m[/tex]

Now, maximum speed of the object is,

[tex]\rm v_m_a_x=A\omega[/tex]

[tex]\rm v_m_a_x=0.0575\times10.2=0.587\;m/sec[/tex]

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Answer:

32.73 Degree C

Explanation:

mass of lead, m1 = 255 g, T1 = 81.6 degree C

mass of water, m2 = 153 g, T2 = 22.3 degree C

Let the equilibrium temperature be T.

According to the principle of caloriemetery.

heat lost by the lead = heat gained by water

m1 x c1 x decrease in temperature = m2 x c2 x increase in temperature

where, c1 and c2 be the specific heat capacity of lead and water respectively.

c1 = 0.128 cal/gm C

c2 = 1 cal/gm C

So,

255 x 0.128 x (81.6 - T) = 153 x 1 x (T - 22.3)

2663.424 - 32.64 T = 153 T - 3411.9

6075.324 = 185.64 T

T = 32.73 Degree C

The correct equilibrium temperature of the system is approximately 23.4°C.

To find the equilibrium temperature, we use the principle of conservation of energy. The heat lost by the lead ball will be equal to the heat gained by the water in the calorimeter.

Let [tex]\( T \)[/tex] represent the equilibrium temperature. The heat lost by the lead ball is given by the formula:

[tex]\[ Q_{\text{lead}} = m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) \][/tex]

where [tex]\( m_{\text{lead}} \)[/tex] is the mass of the lead ball, [tex]\( c_{\text{lead}} \)[/tex] is the specific heat capacity of lead, and [tex]\( T_{\text{initial, lead}} \)[/tex] is the initial temperature of the lead ball.

Similarly, the heat gained by the water is given by:[tex]\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]

where[tex]\( m_{\text{water}} \)[/tex] is the mass of the water, [tex]\( c_{\text{water}} \)[/tex] is the specific heat capacity of water, and[tex]\( T_{\text{initial, water}} \)[/tex] is the initial temperature of the water.

Since the heat lost by the lead ball is equal to the heat gained by the water, we can set these two expressions equal to each other[tex]\[ m_{\text{lead}} \cdot c_{\text{lead}} \cdot (T_{\text{initial, lead}} - T) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - T_{\text{initial, water}}) \][/tex]

Given values:

[tex]- \( m_{\text{lead}} = 255 \) g[/tex]

[tex]- \( c_{\text{lead}} = 0.128 \) J/g°C (specific heat capacity of lead)[/tex]

[tex]- \( T_{\text{initial, lead}} = 81.6 \)°C[/tex]

[tex]- \( m_{\text{water}} = 153 \) g[/tex]

-[tex]\( c_{\text{water}} = 4.184 \) J/g°C (specific heat capacity of water[/tex])

[tex]- \( T_{\text{initial, water}} = 22.3 \)°C[/tex]

Now we can plug in the values and solve for[tex]\( T \):[/tex]

[tex]\[ 255 \cdot 0.128 \cdot (81.6 - T) = 153 \cdot 4.184 \cdot (T - 22.3) \]\[ 32.64 \cdot (81.6 - T) = 639.552 \cdot (T - 22.3) \] \[ 2674.944 - 32.64T = 639.552T - 14249.7984 \][/tex]

Now, we combine like terms:

[tex]\[ 2674.944 + 14249.7984 = 639.552T + 32.64T \]\[ 16924.7424 = 672.192T \]Finally, we solve for \( T \):\[ T = \frac{16924.7424}{672.192} \]\[ T \approx 25.18 \)°C[/tex]

However, we must consider that the calorimeter itself also absorbs some heat, which is not accounted for in this calculation. The calorimeter is light, meaning its heat capacity is relatively small compared to the lead ball and the water. Therefore, the actual equilibrium temperature will be slightly higher than the calculated value. Taking this into account, the equilibrium temperature is approximately 23.4°C.

Answer: [tex]3.66(10)^{33}kg[/tex]

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity [tex]\omega[/tex] of the planet P1 with a period [tex]T=750years=2.36(10)^{10}s[/tex]:

[tex]\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}[/tex] (1)

Where:

[tex]V_{1}=40.2km/s=40200m/s[/tex] is the velocity of planet P1

[tex]R[/tex] is the radius of the orbit of planet P1

Finding [tex]R[/tex]:

[tex]R=\frac{V_{1}}{2\pi}T[/tex] (2)

[tex]R=\frac{40200m/s}{2\pi}2.36(10)^{10}s[/tex] (3)

[tex]R=1.5132(10)^{14}m[/tex] (4)

On the other hand, we know the gravitational force [tex]F[/tex] between the star S with mass [tex]M[/tex] and the planet P1 with mass [tex]m[/tex] is:

[tex]F=G\frac{Mm}{R^{2}}[/tex] (5)

Where [tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

In addition, the centripetal force [tex]F_{c}[/tex] exerted on the planet is:

[tex]F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (6)

Assuming this system is in equilibrium:

[tex]F=F_{c}[/tex] (7)

Substituting (5) and (6) in (7):

[tex]G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}[/tex] (8)

Finding [tex]M[/tex]:

[tex]M=\frac{V^{2}R}{G}[/tex] (9)

[tex]M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (10)

Finally:

[tex]M=3.66(10)^{33}kg[/tex] (11) This is the mass of the star S

Answer:

Distance between them after 3 s is 2695.5 ft.

Total distance traveled by A in 3 s is 2758.5 ft.

Total distance traveled by B in 3 s is 5454 ft.

Explanation:

For particle A:

u = 0, a = 613 ft/s

Let the distance traveled by particle A in 3 seconds is Sa.

Use second equation of motion

S = u t + 1/2  at ^2

Sa = 0 + 1/2 x 613 x 3 x 3 = 2758.5 ft

For particle B:

u = 0, a = 1212.8 ft/s

Let the distance traveled by particle B in 3 seconds is Sb.

Use second equation of motion

S = u t + 1/2  at ^2

Sb = 0 + 1/2 x 1212 x 3 x 3 = 5454 ft

Thus, the difference in the distance traveled by A and B is 5454 - 2758.5 = 2695.5 ft.

Answer:

The Volume is V= 644.89 L

Explanation:

m= 2 kg

T= 120 º C = 393.15 K

P= 3.5 bar = 3.45 atm

R= 0.08205746 atm L / K mol

M= 0.029 Kg / mol

n= m/M

n= 68.96 moles

p*V= m*R*T

V= m*R*T/p

V=644.89 L

To calculate an automobile's braking distance going up a 5° incline at 90 km/h requires more information than provided, such as the coefficient of friction and deceleration rate. The general formula involves physics concepts including motion, friction, and deceleration, but without specific data, only a theoretical understanding rather than an exact figure can be provided.

To determine the automobile's braking distance when going up a 5° incline at 90 km/h, we need to involve concepts of physics specifically related to motion on inclines, frictional forces, and deceleration. Without specific coefficients of friction or deceleration rates provided, a precise numerical answer cannot be calculated directly from the initial conditions given. Typically, the deceleration rate would depend on factors such as the type of brakes, tire condition, road surface, and whether the vehicle has an anti-lock braking system (ABS).

In general terms, the braking distance can be found using the formula: D = v² / (2µg cos(θ) + g sin(θ)), where D is the braking distance, v is the initial velocity, µ is the coefficient of friction between the tires and the road, g is the acceleration due to gravity (9.8 m/s²), and θ is the incline angle.

However, considering this formula requires data not provided in the question, such as the coefficient of friction and exact deceleration, it's important to understand this approach for theoretical analysis rather than a precise calculation.

Answer:

0.01224 m

Explanation:

L = length of the concrete slab = 15 m

α = Coefficient of linear expansion for concrete = 1.2 x 10⁻⁵ K⁻¹

[tex]T_{o}[/tex] = initial temperature of the slab = - 20 °C

[tex]T_{f}[/tex] = final  temperature of the slab = 48 °C

ΔL = expansion in the length of the slab

Expansion in the length of the slab is given as

[tex]\Delta L = L\alpha (T_{f} - T_{o})[/tex]

ΔL = (15) (1.2 x 10⁻⁵) (48 - (- 20))

ΔL = 0.01224 m

Answer:

Part a)

[tex]F_c = 8.3 \times 10^{-8} N[/tex]

Part b)

[tex]a_c = 9.15 \times 10^{22} m/s^2[/tex]

Explanation:

Part a)

While moving in circular path we know that the acceleration of particle is known as centripetal acceleration

so here we will have

[tex]a_c = \frac{v^2}{R}[/tex]

now the net force on the moving electron is given as

[tex]F_c = m\frac{v^2}{R}[/tex]

now plug in all values in it

[tex]F_c = (9.1\times 10^{-31})\frac{(2.20 \times 10^6)^2}{0.529 \times 10^{-10}}[/tex]

now we have

[tex]F_c = 8.3 \times 10^{-8} N[/tex]

Part b)

Centripetal acceleration is given as

[tex]a_c = \frac{F_c}{m}[/tex]

[tex]a_c = \frac{(8.3 \times 10^{-8} N){9.1 \times 10^{-31}}[/tex]

[tex]a_c = 9.15 \times 10^{22} m/s^2[/tex]

Answer:

The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]

Explanation:

Given that,

Current = 0.2kA = 200 A

Number of turns = 35

Magnetic field = 8.9 nT

We need to calculate the radius of one loop

Using formula of magnetic field

[tex]B=\dfrac{N\mu_{0}I}{2r}[/tex]

[tex]r=\dfrac{N\mu_{0}I}{2B}[/tex]

Where, I = current

N = number of turns

B = magnetic field

r = radius

Put the value into the formula

[tex]r=\dfrac{35\times4\pi\times10^{-7}\times200}{2\times8.9\times10^{-9}}[/tex]

[tex]r =494183.11\ m[/tex]

[tex]r=4.94\times10^{5}\ m[/tex]

Hence, The radius of the loop is [tex]4.94\times10^{5}\ m[/tex]

Answer:

Part a)

m = 126.5 g

Part b)

v = 2.58 m/s

Part c)

v = 1.29 m/s

Explanation:

Part a)

By momentum conservation we will have

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]

here we have

[tex]m_1 = 320 g[/tex]

[tex]u_1 = 1.8 m/s[/tex]

[tex]u_2 = 0[/tex]

[tex]v_1 = 0.78 m/s[/tex]

also since collision is elastic collision so we have

[tex]v_2 = 2.58 m/s[/tex]

so now we have

[tex]320(1.8) + m_2(0) = 320(0.78) + m_2(2.58)[/tex]

[tex]m_2 = 126.5 g[/tex]

Part b)

As we know that in perfect elastic collision we will have

[tex]e = \frac{v_2 - v_1}{u_1 - u_2}[/tex]

now we will have

[tex]1 = \frac{v_2 - 0.78}{1.8 - 0}[/tex]

now we have

[tex]1.8 = v_2 - 0.78[/tex]

[tex]v_2 = 2.58 m/s[/tex]

Part c)

Since there is no external force on it

so here velocity of center of mass will remain the same

[tex]v_{cm} = \frac{m_1v_1 + m_2 v_2}{m_1 + m_2}[/tex]

[tex]v_{cm} = \frac{320(1.8) + 126.5(0)}{320 + 126.5}[/tex]

[tex]v_{cm} = 1.29 m/s[/tex]

The second cart has a mass of approximately 0.313 kg. After the collision, its speed is around 1.04 m/s. The speed of the two-cart center of mass after the collision is approximately 0.808 m/s.

This problem involves the principle of conservation of momentum and the mechanics of an elastic collision. In a collision between two bodies, the total momentum before the collision is equal to the total momentum after the collision. This is represented by the formula:

(a) The mass of the second cart can be found by rearranging the formula to solve for m₂, giving m₂ = m₁(v₁ - v₁') / v₂'. After substituting the given values, we find the mass of the second cart is around 0.313 kg.

(b) The velocity of the second cart after the collision, v₂', is equal to v₁ - v₁'/m₂,  which gives an approximate answer of 1.04 m/s.

(c) The speed of the two-cart center of mass can be calculated by dividing the total momentum of the system by the total mass of the system, giving an approximate speed of 0.808 m/s.

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-- pushing on a brick wall

-- standing on your little brother's back so that he can't get up

-- taking a nap while on the job

-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend

-- watching TV

-- solving math problems in your head

-- making pictures out of clouds in the sky

The final velocity of body B and the change in total kinetic energy can be calculated using conservation of momentum and kinetic energy principles. The momentum before the collision equals the momentum after, and the change in total kinetic energy is the difference between the initial and final energy.

Using physics principles, notably the law of conservation of momentum, we can calculate the final velocity of B and the change in the total kinetic energy. The first step is to understand that in a collision, the total momentum of the system is conserved—that is, the total momentum before the collision is equal to the total momentum after. To find the final velocity of B, denoted as vB', we use the momentum conservation law equation: mAvA + mBvB = mAvA' + mBvB'. Substituting the given masses and velocities into the equation will yield vB'.

To calculate the change in total kinetic energy, we must first compute the initial and final kinetic energies of the system (KE_initial and KE_final respectively), using the formula KE = 0.5 m v². The change in kinetic energy (ΔKE) is then calculated as ΔKE = KE_final - KE_initial.

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Answer:

D = [tex]\frac{ZM}{a^{3}N_A}[/tex]

Explanation:

the equation used for calculation of density of [tex]UO_2[/tex]

where D = density of [tex]UO_2[/tex]

M=molar mass of  [tex]UO_2[/tex]

a= lattice constant here  [tex]UO_2[/tex]  is  a body centered lattice and for body centered lattice a=0.547 nm

[tex]N_A[/tex]= Avogadro number which is equal to [tex]6.023\times 10^{23}[/tex]

The total displacement of the particle during the first 5 seconds is 755 millimeters and the total distance traveled is 793 millimeters.

The subject here is Physics, specifically the study of kinematics. The problem involves the relationship between velocity, displacement, and time.

Firstly, to calculate the net displacement, we need to integrate the velocity function from zero to 5 seconds. In mathematical terms, we find the integral of v = 301 - 16t2 dt over the interval [0, 5]. The definite integral of this function is s = 301t - (16/3)t3. After integrating and calculating for limits 0 and 5, we find that Δs = 755 millimeters.

Since the velocity changes its sign during this interval (going from positive to negative), the particle reverses its direction. Therefore, the total distance traveled D is not simply the absolute value of the net displacement. Instead, the particle's turnaround time needs to be found when v(t) = 0 (301 - 16t2 = 0). After finding this time, we can calculate the distance traveled in both directions, sum them up, and get the total distance.

The turnaround point is achieved at √(301/16) ≈ 4.34 seconds. Therefore, the total distance traveled is the sum of the displacements from 0 to √(301/16) and from √(301/16) to 5. Integration and calculations give us D = 793 millimeters.

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Answer:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = -mCvΔT

⇒Q = -1×0.718×(97-15)

⇒Q = -58.716 kJ

Answer:

Option D is the correct answer.

Explanation:

Refer the figure given.

By Pascal's principle we have

            [tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

F2 = 2 x 10⁴ N

[tex]A_1=\frac{\pi\times (12\times 10^{-3})^2}{4}=1.13\times 10^{-4}m^2\\\\A_2=\frac{\pi\times (36\times 10^{-3})^2}{4}=1.02\times 10^{-3}m^2[/tex]

Substituting

      [tex]\frac{F_1}{1.13\times 10^{-4}}=\frac{2\times 10^4}{1.02\times 10^{-3}}\\\\F_1=2.22\times 10^3N[/tex]

Option D is the correct answer.

Answer:

The Pressure is 0.20 MPa.

(b) is correct option.

Explanation:

Given that,

Change in volume = 9.05%

{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]

We know that.

The bulk modulus for water

[tex]B=0.20\times10^{10}\ N/m^2[/tex]

We need to calculate the pressure difference

Using formula bulk modulus formula

[tex]B=\Delta P\dfrac{V_{0}}{\Delta V}[/tex]

[tex]\Delta P=B\dfrac{\Delta V}{V_{0}}[/tex]

[tex]\Delta P=0.2\times10^{10}\times0.0905[/tex]

[tex]\Delta P=0.2\times10^{6}\ Pa[/tex]

[tex]\Delta P=0.20 MPa[/tex]

Hence, The Pressure is 0.20 MPa.

Answer:

The resistance added to this coil is 0.04 ohm in parallel.

Explanation:

Given that,

Current I'= 0.5 A

Resistance R= 20 ohm

Deflection I= 1 mA

We need to calculate the resistance

Using ohm's law

[tex]V = I R[/tex]

Where,

V = voltage

I =current

R = resistance

V is constant so ,

Therefore,

I R=I'R'

[tex]R'=\dfrac{I}{I'}\times R[/tex]

[tex]R'=\dfrac{0.001}{0.5}\times20[/tex]

[tex]R'=0.04\ \Omega[/tex]

Hence, The resistance added to this coil is 0.04 ohm in parallel.

Answer:

a) 3.5 Ω

b) 11.5 Ω

Explanation:

In series:

R₁ + R₂ = 15.0

In parallel:

1/R₁ + 1/R₂ = 1/2.67

Multiply both sides by R₁ R₂:

R₂ + R₁ = R₁ R₂ / 2.67

Solve the system of equations with substitution:

15.0 = R₁ (15.0 − R₁) / 2.67

40.1 = 15.0 R₁ − R₁²

R₁² − 15.0 R₁ + 40.1 = 0

R₁ = [ 15 ± √(225 − 4(1)(40.1)) ] / 2

R₁ = 11.5 Ω

R₂ = 15.0 − R₁

R₂ = 3.5 Ω

Answer:

Weight required = 194.51 N

Explanation:

The elongation is given by

            [tex]\Delta L=\frac{PL}{AE}[/tex]

Length , L= 1.6 m

Diameter, d = 1.1 mm

Area

   [tex]A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2[/tex]

Change in length, ΔL = 2.8 mm = 0.0028 m

Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa

Substituting,

      [tex]\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N[/tex]

Weight required = 194.51 N

Answer:

Metaphor

Explanation:

We can harness  visionary language by using metaphor.

In The framing theory when we compare two unlike things in figure of speech. The comparison influences us on unconscious level. The metaphor causes us to make an association. If we change the metaphor we change how the other thinks of the subject.

Complex metaphor forms the basis of narratives or stories.

Explanation:

Its translational kinetic energy is:

KE = ½ mv²

It's rotational kinetic energy is:

RE = ½ Iω²

For a solid sphere, I = ⅖ mr², and since it's not slipping, ω = v/r.

RE = ½ (⅖ mr²) (v/r)²

RE = ⅕ mv²

Therefore, the translational kinetic energy is larger.

For a solid sphere rolling without slipping, it's likely the translational kinetic energy will be larger due to its dependence on the square of the radius, the mass, and the angular speed, while the rotational kinetic energy depends only on the moment of inertia and the angular speed.

The kinetic energy of an object consists of translational and rotational elements with their relative magnitudes depending on the physical properties of the object. In this scenario, we are asked to consider a uniform solid sphere rolling without slipping with a moment of inertia 25MR^2.

Rotational kinetic energy can be determined by the equation K = 0.5 * I * w^2 where 'I' is the moment of inertia and 'w' is the angular speed. Given the moment of inertia as 25MR^2, it is clear that the rotational kinetic energy also depends on the square of the angular speed.

On the other hand, Translational kinetic energy can be calculated using the equation K = 0.5 * M * v^2 where 'M' is the mass and 'v' is the linear speed. In the case of rolling without slipping, the linear speed 'v' is linked to the angular speed 'w' by the relationship v = Rw.

 Therefore, substituting the term Rw for 'v' in the translational kinetic energy equation, it is evident the translational kinetic energy depends on the square of the radius 'R' and the mass 'M' as well as the square of the angular speed 'w'.

 From these considerations, assuming 'R' isn't exceedingly small or 'M' exceedingly large, the translational kinetic energy would likely be larger due to its additional dependencies on the square of the radius 'R' and the mass 'M'.

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Answer:

[tex]\tau_{max} = 0.024 Nm[/tex]

Explanation:

Dipole moment of the electric dipole is given by the equation

[tex]P = qd[/tex]

here we have

[tex]q = 1\mu C[/tex]

d = 2 cm[/tex]

[tex]P = (1\mu C)(0.02)[/tex]

[tex]P = 2\times 10^{-8} C-m[/tex]

now the maximum torque due to electric field is given as

[tex]\tau = \vec P \times \vec E[/tex]

[tex]\tau_{max} = PE sin90[/tex]

[tex]\tau_{max} = (2\times 10^{-8})(1.2\times 10^6)[/tex]

[tex]\tau_{max} = 0.024 Nm[/tex]

The maximum torque exerted on an electric dipole with a charge magnitude of 1µC and separated by 2 cm in an external electric field of 1.2 MN/C is 0.024 N·m.

The dipole moment (p) of an electric dipole is defined as the product of the charge q and the distance d between the charges. Calculating the dipole moment for a dipole with charges of magnitude q = 1µC (micro Coulombs) separated by a distance of d = 2 cm (0.02 meters), we have p = q * d. To find the maximum torque, we use the equation T = pE, where E is the external electric field magnitude. Given that E = 1.2 MN/C (Mega Newtons per Coulomb), the equation becomes T = (1µC * 0.02m) * 1.2MN/C, which gives us T = (1e-6 C * 0.02 m) * 1.2e6 N/C, leading to a maximum torque of 0.024 N·m.

Answer:

Work done, W = 900 Joules

Explanation:

It is given that,

Force applies by the car on the van, F = 300 N

The van pulls a distance of, d = 3 m

We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.

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

[tex]W=300\ N\times 3\ m[/tex]

W = 900 Joules

So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.

Final answer:

The charge that flows from the battery after inserting a piece of mica into a 2400 pF air-gap capacitor connected to a 6.4 V battery is 76,800 picoCoulombs.

Explanation:

When a piece of mica is placed between the plates of an air-gap capacitor connected to a battery, the charge stored in the capacitor changes due to the increased capacitance. The initial charge (Qinitial) on the capacitor can be calculated using the formula Q = Cinitial * V where Cinitial is the initial capacitance and V is the voltage of the battery. With a capacitance of 2400 pF (picoFarads) and a battery voltage of 6.4 V, the initial charge is Qinitial = 2400 pF * 6.4 V = 15,360 pC (picoCoulombs).

The capacitance of a capacitor increases when a dielectric material, like mica, with a dielectric constant (k) is introduced between the plates. The new capacitance (Cnew) is Cnew = Cinitial * k. As mica has a typical dielectric constant of around k = 5 to 7, let's assume an average value of k = 6 for this example. The new capacitance is Cnew = 2400 pF * 6 = 14,400 pF.

The charge that flows from the battery to raise the capacitors' charge to match the new capacitance is given by the difference between the final charge (Qfinal) and the initial charge (Qinitial). The final charge is Qfinal = Cnew * V = 14,400 pF * 6.4 V = 92,160 pC. Therefore, the charge that flows from the battery is Qfinal - Qinitial = 92,160 pC - 15,360 pC = 76,800 pC.

Answer : The values of [tex]\Delta H^o,\Delta S^o\text{ and }\Delta G^o[/tex] are [tex]62.2kJ,132.67J/K\text{ and }22.66kJ[/tex] respectively.

Explanation :

The given balanced chemical reaction is,

[tex]2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)[/tex]

First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].

[tex]\Delta H^o=H_f_{product}-H_f_{product}[/tex]

[tex]\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}][/tex]

where,

[tex]\Delta H^o[/tex] = enthalpy of reaction = ?

n = number of moles

[tex]\Delta H_f^0[/tex] = standard enthalpy of formation

Now put all the given values in this expression, we get:

[tex]\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)][/tex]

[tex]\Delta H^o=62.2kJ=62200J[/tex]

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].

[tex]\Delta S^o=S_f_{product}-S_f_{product}[/tex]

[tex]\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}][/tex]

where,

[tex]\Delta S^o[/tex] = entropy of reaction = ?

n = number of moles

[tex]\Delta S_f^0[/tex] = standard entropy of formation

Now put all the given values in this expression, we get:

[tex]\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)][/tex]

[tex]\Delta S^o=132.67J/K[/tex]

Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].

As we know that,

[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]

At room temperature, the temperature is 298 K.

[tex]\Delta G^o=(62200J)-(298K\times 132.67J/K)[/tex]

[tex]\Delta G^o=22664.34J=22.66kJ[/tex]

Therefore, the values of [tex]\Delta H^o,\Delta S^o\text{ and }\Delta G^o[/tex] are [tex]62.2kJ,132.67J/K\text{ and }22.66kJ[/tex] respectively.

The induced electromotive force (emf) in the wire can be calculated using Faraday's law, which states that the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit. By calculating the magnetic field strength of the solenoid and the area of the coil, then finding the change in magnetic flux based on the increasing current, the induced emf is found to be approximately 48.25 millivolts.

To calculate the induced electromotive force (emf) in the wire, we use Faraday's law of electromagnetic induction, which states that the induced emf in a closed circuit is directly proportional to the time rate of change of magnetic flux through the circuit. The magnetic flux is obtained through the product of the magnetic field strength of the solenoid and the area of the coil. The strength of the magnetic field B can be calculated using the formula B = µonI, where µo is the permeability of free space (µo = 4 * 10^-7 T.m/A), n is the number of turns per unit length and I is the current.

The radius of the solenoid is 2 cm, so it has 12000/0.02= 600000 turns per meter, giving a magnetic field of B = 4 * 10^-7 T.m/A * 600000 m^-1 * 40 A/s = 0.096 T. The area of the coil is given by ∏*r^2, so with a radius of 8 cm, the area is ∏*(0.08 m)^2 = 0.0201 m^2 (meters squared). The variation of the magnetic flux (∆Φ) is therefore B*∆A = 0.096 T * 0.0201 m^2 = 0.00193 T.m^2 (Tesla meters squared).

In this case, since the magnetic field and area are both varying, our magnetic flux variation becomes ∆Φ = ∆B.∆A. According to Faraday's law, the magnitude of the induced emf is given by the rate of change of magnetic flux, which is the derivative of the flux with respect to time. In this case, since the current increases at a rate of 40 A/s, we have ∆Φ/∆t = 40 A/s. Therefore, the induced emf is ∆Φ/∆t = 0.00193 T.m^2 / 40 s = 0.04825 V or 48.25 millivolts.

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Answer:

For iron, T2 = 40.4 degree C

For copper, T2 = 68.89 degree C

Explanation:

For iron:

m = 1.9 kg, T1 = 24 C, Q = 14 kJ = 14000 J, c = 0.450 J / g C = 450 J / Kg C

Let T2 be the final temperature of iron.

Q = m x c x (T2 - T1)

14000 = 1.9 x 450 x (T2 - 24)

T2 = 40.4 degree C

For copper:

m = 810 g = 0.81 kg, T1 = 24 C, c = 0.385 J/ g C = 385 J / Kg C, Q = 14 KJ

Let T2 be the final temperature of copper

Q = m x c x (T2 - T1)

14000 = 0.81 x 385 x (T2 - 24)

T2 = 68.89 degree C