Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generates electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.
Part (a) Calculate the power in this flow in watts.

Answer:

B = 1.353 x 10⁻³ T

Explanation:

The Magnetic field within a toroid is given by

B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.

To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,

r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.

b = 25.1 + 3 = 28.1 cm

a = 25.1 cm

r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m

B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266

B = 1.353 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section of a given toroid, calculated using Ampere's Law, is approximately 0.00269 T.

The magnetic field B produced by a current in a toroidal solenoid is described by Ampere's Law. The formula to calculate magnetic field (B) in a toroid is B = μ₀IN / (2πr), where:

Substituting these values in the formula, we get:

B = (4π x 10⁻⁷ T m/A) * (600) * (3 A) / (2π * 0.266 m) = 0.00269 T.

So the strength of the magnetic field at the center of the square cross section is approximately 0.00269 T.

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Answer:

11 radians or 1.7507 revolutions

9 rad/s

7.27565 revolutions

Explanation:

[tex]\omega_f[/tex] = Final angular velocity

[tex]\omega_i[/tex] = Initial angular velocity

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

[tex]\theta[/tex] = Angle of rotation

t = Time taken

Equation of rotational motion

[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\times 2+\frac{1}{2}\times 3.5\times 2^2\\\Rightarrow \theta=11\ rad=\frac{11}{2\pi}=1.7507\ rev[/tex]

Angle the wheel rotates in the given time is 11 radians or 1.7507 revolutions

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=2+3.5\times 2\\\Rightarrow \omega_f=9\ rad/s[/tex]

The angular speed of the wheel at the given time is 9 rad/s

[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{(9\times 2)^2-2^2}{2\times 3.5}\\\Rightarrow \theta=45.71428\ rad=\frac{45.71428}{2\pi}=7.27565\ rev[/tex]

The number of revolutions if the final angular speed doubles is 7.27565 revolutions

This question involves the concepts of the equations of motion for angular motion.

(a) The wheel rotates through the angle "11 rad" (OR) "1.75 rev".

(b) The angular speed at t = 2 s is "9 rad/s".

(c) The angular displacement for double speed is "45.71 rad" (OR) "7.28 rev".

(a)

The angular displacement can be found using the second equation of motion for angular motion.

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

where,

θ = angular displacement = ?

ωi = initial angular speed = 2 rad/s

t = time interval = 2 s

α = angular acceleration = 3.5 rad/s²

Therefore,

[tex]\theta = (2\ rad/s)(2\ s)+\frac{1}{2}(3.5\ rad/s^2)(2\ s)^2\\\theta = 4\ rad\ +\ 7\ rad[/tex]

θ = 11 rad

[tex]\theta = (11\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]

θ = 1.75 rev

(b)

The angular speed can be found using the first equation of motion for angular motion.

[tex]\omega_f = \omega_i+\alpha t\\\omega_f = 2\ rad/s + (3.5\ rad/s^2)(2\ s)\\[/tex]

ωf = 9 rad/s

(c)

The angular displacement can be found using the third equation of motion for angular motion after we double the value of ωf.

[tex]2\alpha \theta = \omega_f^2-\omega_i^2\\2(3.5\ rad/s^2)\theta = (18\ rad/s)^2-(2\ rad/s)^2\\\\\theta = \frac{320\ rad^2/s^2}{7\ rad/s^2}\\\\[/tex]

θ = 45.71 rad

[tex]\theta = (45.71\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]

θ = 7.28 rev

Learn more about the angular motion here:

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The attached picture shows the angular equations of motion.

Answer:

amount of work done is[tex] W  =  \frac{-4}{3}[/tex]      

Explanation:

Formula for work done by force field

[tex]W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt[/tex]

where

r(t) is parametrization of line

as it is straight line so

[tex]r(t) = 1- t) r_o + tr_1   0 \leq t \leq 1[/tex]

thus,

r(t) = (1-t)(-1,1) + t(3,-3)

    = (-1+t,1-t) + (3t - 3t)

   = (-1+t +3t, 1-t-3t)

r(t) = (4t -1, 1- 4t)

r'(t) = (4,-4)

putting value in above integral

[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \int_{0}^{1} (-(4t -1)^2 , 2-8t).(4,-4) dt[/tex]

                                                  [tex]= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt[/tex]

                                                  [tex] =\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt[/tex]

                                                   [tex]= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}[/tex]

[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \frac{-4}{3}[/tex]                                                        

To find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1) to (3,-3), we compute the line integral of the dot product of the force and displacement vectors along the path. Performing the integrations using the limits set by the line's endpoints gives us the work done.

The task involves physics concepts in vector calculus. Specifically, we're looking at the work done by a force vector over a path in two-dimensional space. Here, the force is described by the vector function ​F = xyi +(y-x)j. We need to calculate the work done by this force as it moves along a straight line path from ​ (-1,1) to (3,-3).

Work done by a force in moving an object is given by the line integral of ​force · displacement. In mathematical terms, this definition reads as W = ∫F.dr. The computation requires us to take the dot product of the force and differential displacement vectors along the path.

The direction of displacement as we go from (-1,1) to (3,-3) is simply the path's tangent. Therefore, dr = dx i + dy j, where (dx,dy) is the differential displacement vector along the line. Since the line is straight, (dx, dy) = (3 - (-1), -3 - 1) = (4, -4) up to a constant. Then dot product becomes F · dr = (xy - (y-x))*(4).

Now, we take the line integral of this product over the path. The limits of integration, derived from the line's endpoints, are x = -1 to x = 3. We carry out the integration and simplify to get the work done by the force over the straight line. Note that the straight-line path simplifies the calculation. For paths with more complex shapes or forces that vary nonlinearly with displacement, such integrations may require specialist mathematical techniques.

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To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.

By definition we know that velocity is defined as the change of position due to time, therefore

[tex]V = \frac{d}{t}[/tex]

Where,

d = Distance

t = Time

Speed can also be expressed in vector form through its components [tex]V_x[/tex] and [tex]V_y[/tex]

In the case of the horizontal component X, we have to

[tex]V_x = \frac{d}{t}[/tex]

Here d means the horizontal displacement, then

[tex]t = \frac{d}{V_x}[/tex]

[tex]t = \frac{67}{V_x}[/tex]

At the same time we have that the vertical component of the velocity is

[tex]V_y = gt[/tex]

Here,

g = Gravity

Therefore using the relation previously found we have that

[tex]V_y = g \frac{67}{V_x}[/tex]

The relationship between the two velocities and the angle can be expressed through the Tangent, therefore

[tex]tan\theta = \frac{V_y}{V_x}[/tex]

[tex]tan \theta = \frac{g \frac{67}{V_x} }{V_x}[/tex]

[tex]tan 3 = \frac{9.8\frac{67}{V_x} }{V_x}[/tex]

[tex]tan 3 = \frac{9.8*67}{V_x^2}[/tex]

[tex]V_x^2 = \frac{9.8*67}{tan 3}[/tex]

[tex]V_x= \sqrt{ \frac{9.8*67}{tan 3}}[/tex]

[tex]V_x = 111.93m/s \hat{i}[/tex]

This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,

Then [tex]V_y,[/tex]

[tex]V_y = g \frac{67}{V_x}[/tex]

[tex]V_y = 9.8*\frac{67}{111.93}[/tex]

[tex]V_y = 5.866m/s\hat{j}[/tex]

[tex]|\vec{V}| = \sqrt{111.93^2+5.866^2}[/tex]

[tex]|\vec{V}| = 112.084m/s[/tex]

Final answer:

To calculate the initial speed of the arrow in a projectile motion problem, one needs to find the time of flight using the horizontal range and the final impact angle and then solve for the initial horizontal and vertical velocity components.

Explanation:

To determine the initial speed of the arrow shot from the bow, we'll use the principle of projectile motion. First, we need to find the time of flight. The arrow makes a 3.00° angle with the ground upon impact, and since it was shot parallel to the ground, it maintains a constant horizontal velocity throughout its flight. So, we can calculate the initial horizontal speed using this final impact angle.

The horizontal range (R) is given by R = v0x * t, where v0x is the initial horizontal velocity and t is the time of flight. The arrow landed 67.0 m away, so R is 67.0 m.

To find t, we will use the vertical motion equation under constant acceleration due to gravity (g = 9.8 m/s2). We know the final vertical velocity component (vfy) can be related to the initial vertical velocity component (v0y = 0 since it was shot parallel to the ground) by vfy = v0y + g*t. We can also calculate vfy using the impact angle: vfy = tan(impact angle) * v0x. Combining these, we can solve for t and subsequently for the initial speed v0.

Answer:

Only option A is correct. Beaker A has lower kinetic energy than beaker B.

Explanation:

Step 1: Data given

Beaker 1 has a volume of 100 mL at 25 °C

Beaker B has a volume of 100 mL at 60 °C

Thermal energy = m*c*T

Thermal energy beaker A = 100 grams*4.184 * 25°C

Thermal energy beaker B = 100 grams *4.184*60°C

⇒ Since both beakers contain the same amount of water, the thermal energy depends on the temperature.

Since beaker B has a higher temperature, it has a higher thermal energy than beaker A

When we heat a substance, its temperature rises and causes an increase in the kinetic energy of its constituent molecules. Temperature is, in fact, a measure of the kinetic energy of molecules.

This means beaker B has a higher kinetic energy than beaker A

Potential energy doesn't depend on temperature. this means the potential energy of beaker A and beaker B is the same.

a. Beaker A has lower kinetic energy than beaker B. This is correct.

b. Beaker A has higher thermal energy than beaker B. This is false.

c. Beaker A has higher potential energy than beaker B. This is false.

d. Beaker A has lower potential energy than beaker B. This is false

e. Beaker A has higher kinetic energy than beaker B. This is false.

Beaker A has lower kinetic energy than Beaker B because the temperature of a substance is directly proportional to its average kinetic energy. Thus, the water in Beaker B, which is at a higher temperature, has higher kinetic energy than the cooler water in Beaker A.

The correct answer to your question is: a. Beaker A has lower kinetic energy than Beaker B. This is because the temperature of a substance is directly proportional to its average kinetic energy. Kinetic energy refers to the energy of movement, so in this case, it's the energy of the water molecules moving within each beaker. In Beaker B, the water is at a higher temperature, meaning the water molecules are moving more quickly, and therefore have higher kinetic energy. On the other hand, in Beaker A, the water is at a lower temperature, so the water molecules are moving more slowly, which means they have lower kinetic energy.

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When a doubled rope, i.e. two pieces of rope used together, is used, it stretches half as much as a single rope would under the same load. Correspondingly, a rope's stretch is directly proportional to its length. So, it will stretch twice as much when its length is doubled.

The stretchiness of a rope under a load is a property called elasticity. When there are two ropes, they share the load between them. Therefore, each carries only half the load. Hence, a doubled rope would stretch half as much as a single rope when supporting the same weight. In this case, the single rope stretches by 20 cm under a load. Thus, the doubled rope would stretch by 10 cm.

Similarly, the amount a rope stretches is proportional to its length. If a 10 m rope stretches 60 cm under a certain load, a 20 m rope -- twice as long -- would stretch twice as much under the same load, or 120 cm.

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In the context of elasticity in physics, a doubled rope that halves each individual rope's load would stretch half as much. Therefore, if a 10m rope stretches by 20cm, a doubled rope would stretch by 10cm. If a 10m rope stretches by 60cm, a 20m rope under the same load would stretch twice as much, or 120cm.

These questions refer to the concepts of strain and stress in physics, specifically within the context of elasticity. Strain refers to the deformation of a body due to the applied stress, in this case, the length of rope extending under the weight of the climber.

a. If a 10m climbing rope stretches by 20cm under a certain load, its strain is 0.02 (20cm/10m). When the load is shared by two ropes, each rope bears half of the load. Hence, each rope is likely to stretch half as much. So, the doubled rope would stretch by 10cm.

b. Stretching is directly proportional to the length of the rope. Therefore, if a 10m rope stretches by 60cm under a load, a 20m rope under the same load would stretch twice as much, which is 120cm or 1.2m.

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Answer:

672.29 W/m²

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

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

I = Intensity of light = 1200 W/m²

[tex]E_m[/tex] = Maximum value electric field

Intensity of light is given by

[tex]I=\frac{1}{2}\epsilon_0cE_m^2\\\Rightarrow E_m=\sqrt{\frac{2I}{\epsilon_0c}}\\\Rightarrow E_m=\sqrt{\frac{2\times 1200}{8.85\times 10^{-12}\times 3\times 10^8}}\\\Rightarrow E_m=950.765\ N/C[/tex]

RMS value

[tex]E_r=\frac{E_m}{\sqrt2}\\\Rightarrow E_r=\frac{950.765}{\sqrt2}\\\Rightarrow E_r=672.29\ W/m^2[/tex]

The approximate magnitude of the electric field in the sunlight is 672.29 W/m²

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, [tex]V_{Hall} = 20\ mV = 0.02\ V[/tex]

Magnetic Field, B = 0.10 T

Hall emf, [tex]V'_{Hall} = 69\ mV = 0.069\ V[/tex]

Now,

Drift velocity, [tex]v_{d} = \frac{V_{Hall}}{B}[/tex]

[tex]v_{d} = \frac{0.02}{0.10} = 0.2\ m/s[/tex]

Now, the expression for the electric field is given by:

[tex]E_{Hall} = Bv_{d}sin\theta[/tex]                            (1)

And

[tex]E_{Hall} = V_{Hall}d[/tex]

Thus eqn (1) becomes

[tex]V_{Hall}d = dBv_{d}sin\theta[/tex]

where

d = distance

[tex]B = \frac{V_{Hall}}{v_{d}sin\theta}[/tex]                      (2)

(a) When [tex]\theta = 90^{\circ}[/tex]

[tex]B = \frac{0.069}{0.2\times sin90} = 0.345\ T[/tex]

(b) When [tex]\theta = 60^{\circ}[/tex]

[tex]B = \frac{0.069}{0.2\times sin60} = 0.398\ T[/tex]

Answer:

A. increase

Explanation:

Stirling cycle having four processes

1.Two processes are constant temperature processes

2.Two processes are constant volume processes

The efficiency of Stirling cycle is same as the efficiency of Carnot cycle.

The efficiency of Stirling cycle given as

[tex]\eta=1-\dfrac{T_L}{T_H}[/tex]

[tex]T_L[/tex]=Lower temperature

[tex]T_H[/tex]=Upper temperature

If we increase then upper temperature while the lower temperature is constant then the efficiency of Stirling cycle will increase because the ratio of lower and upper temperature will decreases.

Therefore answer is

A. increase

Final answer:

The angular acceleration of the yo-yo is 850 rad/s², the angular velocity after 0.750 s is 637.5 rad/s, and the tangential acceleration of a point on its edge is 28.05 m/s².

Explanation:

To solve the yo-yo problem:
(a) The angular acceleration (α) can be found by using the linear acceleration (a) of the yo-yo and the radius (r) of its center shaft. The formula is α = a/r. Given that the yo-yo accelerates at 1.70 m/s2 and the radius of the center shaft is 0.200 cm (0.002 m), the angular acceleration is α = 1.70 m/s2 / 0.002 m = 850 rad/s2.
(b) The angular velocity (ω) after a certain time can be calculated using the formula ω = αt, assuming it starts from rest. So after 0.750 s, the angular velocity is ω = 850 rad/s2 * 0.750 s = 637.5 rad/s.
(c) To find the tangential acceleration (at) of a point on the edge of the yo-yo, with an outside radius of 3.30 cm (0.033 m), we use the formula at = α * R. Therefore, the tangential acceleration at the edge is at = 850 rad/s2 * 0.033 m = 28.05 m/s2.

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

(c) e = NBAω Sinωt

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

(A) A device that converts heat into work with 100% efficiency

It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.

The concepts required to solve this problem are the thermodynamic expressions of expansion and linear expansion.

In mathematical terms the dilation of a body can be expressed as

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

Where,

[tex]L_0 =[/tex] Initial Length

[tex]\alpha =[/tex] Thermal coefficient of linear expansion

[tex]\Delta T =[/tex] Change in Length

Our values are given as

[tex]L_0 = 300m[/tex]

[tex]T_f = 25\°C[/tex]

[tex]T_i = 2\°C[/tex]

[tex]\alpha = 12*10^{-6}C^{-1} (Iron)[/tex]

Replacing at the equation we have,

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

[tex]\Delta L = (300)(12*10^{{-6})(25-2)[/tex]

[tex]\Delta L = 0.0828m[/tex]

Therefore the change in the height is 8.28cm

The Eiffel Tower's height is subject to changes due to thermal expansion. In fact, it's around 8.3 cm taller in July than in January, with the temperature rise contributing to this change.

The height change of the Eiffel Tower due to thermal expansion can be estimated using the linear expansion formula: ΔL = α*L*ΔT. Here, ΔL is the change in length, α is the thermal expansion coefficient, L is the original length, and ΔT is the change in temperature. Wrought iron has an expansion coefficient of approximately 12*10^-6 °C^-1. The height of the Eiffel Tower is approximately 300 m. The difference in average temperatures is approximately 23°C (25°C-2°C).

Let's plug these values into the formula: ΔL = 12*10^-6 °C^-1 * 300m * 23°C, which gives us a change in height of approximately 0.083m, or 8.3 cm. Therefore, the Eiffel Tower is an estimated 8.3 cm taller in July than it is in January, due to the effects of thermal expansion.

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Answer:

  [tex]T_1 =677224.40\ N[/tex]

Explanation:

given,

mass of the both ball = 5 Kg

length of rod = 2 L            

where L = 0.55 m            

angular speed = 45.6 rev/s

ω = 45.6 x 2 π                      

ω = 286.51 rad/s                

v₁ = r₁ ω₁                        

v₁ =0.55 x 286.51 = 157.58 m/s

v₂ = r₂ ω₂                                

v₂ = 1.10 x 286.51 = 315.161 m/s

finding tension on the first half of the rod

r₁ = 0.55  r₂ = 2 x r₁ = 1.10

  [tex]T_1 = m (\dfrac{v_1^2}{r_1}+\dfrac{v_2^2}{r_2})[/tex]

  [tex]T_1 = 5 (\dfrac{157.58^2}{0.55_1}+\dfrac{315.161^2}{1.1})[/tex]

  [tex]T_1 =677224.40\ N[/tex]

Answer:

K= 290.28 N/m

Explanation:

Given: mass= 85 kg

the time taken to reach point two more times in 6.8 s.

2×t= 6.8 sec

t= 6.8/2= 3.4 sec

then, the time period for oscillation is

[tex]t= 2\pi\sqrt{\frac{m}{k} }[/tex]

Here K= spring constant

m= mass of jumper

⇒[tex]K= \frac{4\pi^2m}{t^2}[/tex]

now plugging the values we get

[tex]K= \frac{4\pi^2\times85}{3.4^2}[/tex]

K= 290.28 N/m

Answer:

a. The stored electric field energy can be greater than the stored magnetic field energy.

b. As the capacitor is discharging, the current is increasing.

d. The stored electric field energy can be less than the stored magnetic field energy.

e. The stored electric field energy can be equal to the stored magnetic field energy.

Explanation:

We know that total energy of capacitor and inductor system given as

[tex]TE=\dfrac{LI^2}{2}+\dfrac{CV^2}{2}[/tex]

L=Inductance  ,I =current

C=Capacitance  ,V=Voltage

The total energy of the system will be remain conserve.If energy in the inductor increases then the energy in the capacitor will decrease and vice -versa.

The values of stored energy in the capacitor and inductor can be equal ,greater or less than to each other.But the summation will be constant always.

Therefore the following option are correct

a ,b , d , e

In a series LC circuit, energy oscillates between the capacitor and inductor. Statements a, b, d, and e are correct as there are moments when the electric field energy can be greater than, less than, or equal to the magnetic field energy. Statement c is incorrect because the current decreases as the capacitor charges.

When an inductor is connected in series with a fully charged capacitor, the behavior of the system can be analyzed by considering the energy stored in both components.

Overall, the series LC circuit exhibits an oscillatory energy transfer between the electric field of the capacitor and the magnetic field of the inductor.

Answer:A

Explanation:

Given

Two ponies are at Extreme end of turntable with mass m

suppose turntable is moving with angular velocity [tex]\omega [/tex]

Moment of inertia of Turntable and two ponies

[tex]I=I_o+2mr_0^2[/tex]

let say at any time t they are at a distance of r from center such [tex]r<r_0[/tex]

Moment of inertia at that instant

[tex]I'=I_0+2mr'^2[/tex]

[tex]I'<I_0[/tex]

conserving angular momentum as net Torque is zero

[tex]I\omega =I'\omega '[/tex]

[tex]\omega '=\frac{I}{I'}\times \omega [/tex]

[tex]\omega '>\omega [/tex]

Thus we can say that angular velocity increases as they move towards each other.        

The angular speed of the turntable increases as the ponies walk towards each other due to conservation of angular momentum. The angular momentum remains constant as there are no external forces acting upon the system.

As the two ponies of equal mass walk toward each other across the turntable, the angular speed of the system increases. This is because the moment of inertia of the system decreases as the ponies move closer to the center, and according to the principle of conservation of angular momentum, a decrease in the moment of inertia must be accompanied by an increase in angular speed to keep the system's angular momentum constant.

In this scenario, yes, the angular momentum of the system is conserved. There are no external forces acting on the system, thus the sum of the angular momenta of the ponies and the turntable stays the same.

This principle can be exemplified by the behavior of figure skaters as they increase their rotation speed by bringing their arms in closer to the body, decreasing their moment of inertia and consequently increasing their angular velocity, to conserve their angular momentum.

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Answer:

The hill should be not less than 0.625 m high

Explanation:

This problem can be solved by using the principle of conservation of mechanical energy. In the absence of friction, the total mechanical energy is conserved. That means that

[tex]E_m=U+K[/tex] is constant, being U the potential energy and K the kinetic energy

[tex]U=mgh[/tex]

[tex]K=\frac{mv^2}{2}[/tex]

When the car is in the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

[tex]mgh=\frac{mv^2}{2}[/tex]

We can solve for h:

[tex]h=\frac{v^2}{2g}=\frac{3.5^2}{2(9.8)}=0.625m[/tex]

The hill should be not less than 0.625 m high

The hill should have a height of 0.625 m and above.

This is defined as the measurement of the vertical position of a body.

Total mechanical energy = Potential energy + kinetic energy.

Potential energy = mgh

Kinetic energy = 1/2mv²

We can infer that:

mgh = 1/2mv²

h = v² / 2g

= (3.5)² / 2(9.8)

= 0.625m.

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Answer:

d

Explanation:

Nuclear reactions unlike chemical reactions do not involve electrons or bonds.The main use for the reaction is neutrons. In nuclear there is only turning the element into another element and releasing a large amount of  energy. So catalyst are also redundant in nuclear reactions.  

Therefore, points III and 3 are true, and rest are all wrong. therefore option d is correct

In a nuclear reaction, elements are converted from one to another and the reaction rates are typically not affected by catalysts. Characteristics such as electrons in atomic orbitals being involved in breaking and forming bonds, and atoms being rearranged by breaking and forming of chemical bonds, pertain to chemical reactions, not nuclear ones.

The correct characteristics of a nuclear reaction are options II and III. In a nuclear reaction, elements are indeed converted from one to another (II). This happens as the structure of an atom's nucleus changes. Furthermore, nuclear reaction rates are typically not affected by catalysts (III). This contrasts with chemical reactions, where catalysts can significantly speed up reaction rates. As for options I and IV, they are characteristic of chemical reactions, not nuclear ones. In chemical reactions, electrons in atomic orbitals are involved in the breaking and forming of bonds (I), and atoms are rearranged by the breaking and forming of chemical bonds (IV). Therefore, the answer is (d) II and III.

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Answer:

ρ = 1469  kg/m³

Explanation:

given,

mass of statue = 0.4 Kg

density of statue = 8 x 10³ kg/m³

tension in the string = 3.2 N

density of the fluid = ?

Volume of the statue

[tex]V = \dfrac{0.4}{8\times 10^3}[/tex]

V = 5 x 10⁻⁵ m³

W = ρ g V

W = ρ x 9.8 x 5 x 10⁻⁵

now, tension on the string will be equal to

T = mg - W

3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵

ρ x 9.8 x 5 x 10⁻⁵ = 0.72

ρ = 1469  kg/m³

Final answer:

The correct answer to the question is (c) The force of gravity acting on the ball. Once the golf ball is in flight and air resistance is neglected, gravity is the only force acting upon it, influencing its parabolic trajectory.

Explanation:

The question asks which forces act on a golf ball while it is in flight, given that air resistance is neglected. The correct option in this case is (c) The force of gravity acting on the ball. When a golf ball is hit and is flying through the air, the force of the golf club is no longer acting on it once it has left contact with the club. There is no force of the ball moving forward through the air; this is a description of motion, not a force. Therefore, the primary force acting on the golf ball once it is in the air is the force of gravity, which is the gravitational pull exerted by the Earth on the ball, causing it to follow a parabolic trajectory. The gravity is also responsible for the ball's downward acceleration at all points in its flight, as it is the only force acting on the ball if we ignore air resistance.

Final answer:

When a golf ball is hit and flying through the air, with air resistance neglected, the only force acting on it is gravity. The force from the golf club is present only during the impact, and the motion of the ball through the air is not a force but the result of the initial momentum given to the ball.

Explanation:

The question addresses the forces acting on a golf ball after it is hit by a golf club. Considering air resistance is neglected, the correct answer is (c) the force of gravity acting on the ball. Once the ball is in flight, the only force acting on it is the gravitational pull of the Earth, which affects the motion of the ball, causing it to follow a parabolic trajectory until it lands. The force of the golf club is only present during the brief contact with the ball, and after that, it doesn't affect the ball's flight. The idea of the force of the ball moving forward through the air is not a force but rather a consequence of the change in the momentum of the golf ball when the club strikes it, resulting in the velocity of the golf ball as it leaves the club.

To solve this problem it is necessary to apply the concepts related to the kinematic equations of motion.

By definition we know that speed can be expressed as

[tex]v= \frac{d}{t}[/tex]

Where,

t = time

d = distance

For this specific case we know that the speed traveled corresponds to the round trip, therefore

[tex]v = \frac{2d}{t}[/tex]

Re-arrange to find t (The time taken to here the echo)

[tex]t = \frac{2d}{v}[/tex]

Replacing with our values we have that the distance is equal to 65m

[tex]t = \frac{2*65}{1530}[/tex]

[tex]t = 0.085s[/tex]

Therefore the time before the dolphin hears the echoes of the clicks is 0.085s

To determine how long it takes for particles to settle at the bottom of a container, principles of sedimentation in physics are applied. The calculation considers gravitational forces, buoyancy, and drag forces. However, exact time estimation requires additional specific information on fluid characteristics.

The question revolves around the concept of sedimentation, which is a physical process where particulate matter settles down at the bottom of a container due to gravity. To calculate how long it takes for spherical particles with a diameter of 2.5 μm (micrometers) and a mass of 1.9 x 10⁻¹⁴ kg to settle at the bottom of a 15-cm-tall container, we need to understand the principles of particle sedimentation in fluids, a topic studied in physics. This involves the forces acting on the particles, including gravitational forces, buoyancy, and drag forces, along with the properties of the fluid and the particles themselves. However, without a specific formula or further details on the characteristics of the fluid (e.g., viscosity) and assuming we ignore any air resistance or buoyancy effects, it's difficult to provide an exact time for sedimentation.

Therefore, it takes approximately [tex]\(4657.76\)[/tex] seconds for all the particles to settle to the bottom of the container.

To find the time it takes for all the particles to settle to the bottom of the container, we can use Stokes' law, which relates the settling velocity of particles to their size and density.

Stokes' law states: [tex]\( v = \frac{{2gr^2(\rho_p - \rho)}}{{9\eta}} \),[/tex] where:

-  v  is the settling velocity of the particle,

-  g  is the acceleration due to gravity [tex](\( 9.8 \, \text{m/s}^2 \)),[/tex]

-  r  is the radius of the particle [tex](\( 1.25 \times 10^{-6} \, \text{m} \)),[/tex]

- [tex]\( \rho_p \)[/tex]is the density of the particle[tex](\( \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3} = \frac{1.9 \times 10^{-14}}{\frac{4}{3}\pi(1.25 \times 10^{-6})^3} \)),[/tex]

- [tex]\( \rho \)[/tex] is the density of the fluid [tex](\( \rho = 1.2 \, \text{kg/m}^3 \)),[/tex] and

- [tex]\( \eta \)[/tex] is the viscosity of the fluid[tex](\( \eta = 1.81 \times 10^{-5} \, \text{Pa s} \)).[/tex]

After substituting the values into Stokes' law and solving for the settling velocity, we found it to be approximately[tex]\(3.22 \times 10^{-5} \, \text{m/s}\).[/tex]

Then, using the formula [tex]\(t = \frac{h}{v}\), where \(h\)[/tex]is the height of the container (0.15 m), we calculated the time it takes for the particles to settle to the bottom:

[tex]\[ t \approx 4657.76 \, \text{s} \][/tex]

Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = [tex]A_{y}[/tex] / A

    Aₓ = A cos 160 = 2 cos 160

    [tex]A_{y}[/tex]  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    [tex]A_{y}[/tex]  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = [tex]B_{y}[/tex] / B

    Bₓ = B cos 260

     [tex]B_{y}[/tex]  = B sin 260

    Bₓ = 4 cos 260

     [tex]B_{y}[/tex]  = 4 sin 260

     Bₓ = -0.6946mi

     [tex]B_{y}[/tex]  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = [tex]C_{y}[/tex] / C

     Cₓ = C cos 15

     [tex]C_{y}[/tex] = C sin15

     Cₓ = 3 cos 15

    [tex]C_{y}[/tex] = 3 sin 15

     Cₓ = 2,898 mi

    [tex]C_{y}[/tex] = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = [tex]A_{y}[/tex] +  [tex]B_{y}[/tex] + [tex]C_{y}[/tex]

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

Answer:

The potential energy in the water will be 0.0769 mJ

Explanation:

We know that energy stored in the capacitor due to charge is given by [tex]U=\frac{Q^2}{2C}[/tex]

From the relation we can see that potential energy is inversely proportional to the capacitance

And we also know that capacitance is directly proportional to the dielectric constant

So the new potential energy will be [tex]=\frac{6}{78}=0.0769mJ[/tex]

So the potential energy in the water will be 0.0769 mJ

Final answer:

When a capacitor filled with air and charged is filled with water without discharging its plates, the energy stored increases due to the higher dielectric constant of water. The formula used is U' = K_water * U, leading to the capacitor storing 468 mJ of energy after being filled with water.

Explanation:

The energy stored in a capacitor is given by the formula U = ½ CV^2, where U is the stored energy, C is the capacitance and V is the voltage across the capacitor. When a dielectric is introduced between the plates of a capacitor that is charged but disconnected from any voltage supply, the capacitance of the capacitor increases by a factor equal to the dielectric constant of the material inserted, K. Since the charge Q remains constant, and since C increases while U is directly proportional to C, the energy stored in the capacitor will increase as well.

For this particular question, the initial energy is 6 mJ and the dielectric constant for water is 78. After filling the capacitor with water, the new energy stored U' can be found using the dielectric constant. However, since dielectric constant for air is approximately 1, and capacitance is directly proportional to dielectric constant, we can simply multiply the original energy by the dielectric constant of water to find the energy stored after it is filled.

U' = K_water * U  

  = 78 * 6 mJ  

  = 468 mJ

Therefore, the energy the capacitor continues to store after it is filled with water is 468 mJ.

Answer:b. In your hand

Explanation: The objects from an outer perspective are moving at 90km/h all of them, even if there was a fly flying around in the train. Then if there is no change of speed of the train the moment the ball leaves the hand its relative speed with respect to the hand is zero (0), then it will land in the hand again. So the only way for it not to happen is either you move the hand or the train changes its speed, therefore the relative speed of the

ball with respect to the hand would be differente from zero and further information would be required, but since it is not stated that neither the train accelerates nor the hand is going to be moved, we can affirm the ball will land in the hand.

Answer:

The velocity of the goalie = 0.156 m/s

The velocity of the puck is 34.85 m/s, but in the reverse direction (-34.85 m/s)

Explanation:

The total momentum before equals the totam momentum after the collision for objects with mass m1 and m2  

m1*vi1 + m2 vi2 = m1 vf1 + m2 vf2

⇒ with vi = the initial velocity

⇒ with vf = the final velocity

initial momentum = final momentum  

An elastic collision is one in which the total kinetic energy of the two colliding objects is the same before and after the collision. For an elastic collision, kinetic energy is conserved. That is:  

0.5 m1 vi1² + 0.5 m2 vi2² = 0.5 m1 vf1² + 0.5 m2 vf2²  

Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:  

vf1 = [(m1 - m2) vi1 + 2m2 vi2]/[m1 + m2]  

vf2 = [2m1vi1 − (m1 - m2) vi2]/[m1 + m2]  

⇒ with m1 = the mass of the goalie = 65 kg

⇒ with m2 = the mass of the puck = 0.145 kg

⇒ with vi1 = the initial velocity of the goalie 0 m/s

⇒ with vi2 = the initial velocity of the puck = 35 m/s

⇒ with vf1 = The final velocity of the goalie

   => vf1 = [(65 - 0.145)0 + 2(0.145)35]/[65.145]  

=  0.1558  ≈ 0.156 m/s. Since it's positve this is in the direction of the puck

=> vf2 = [0 − (65 - 0.145)35]/[65.145] = -34.85 m/s

The velocity of the puck is 34.85 m/s, in the reverse direction. (Has a negative sign)

Answer:

[tex]a =29.54\ m/s^2[/tex]

Explanation:

given,

radius of CD player = 12 cm

assume rate of acceleration = 100 rad/s²

times = 0.15 s

now,

tangential acceleration  

  [tex]a_t = \alpha r[/tex]

  [tex]a_t = 100 \times 0.12[/tex]

  [tex]a_t = 12 m/s^2[/tex]

now using equation

v = v₀ + a_t x t

v =0+ 12 x 0.15

v = 1.8 m/s

now, radial acceleration

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

[tex]a_r = \dfrac{1.8^2}{0.12}[/tex]

[tex]a_r =27\ m.s^2[/tex]

now

acceleration

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

[tex]a = \sqrt{27^2+12^2}[/tex]

[tex]a =29.54\ m/s^2[/tex]

To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as

[tex]\Delta \theta = 1.22\frac{\lambda}{d}[/tex]

Where,

\lambda = Wavelength

d = Optical Diameter

[tex]\theta =[/tex] Angular resolution

In turn you can calculate the angle through the diameter and the arc length, that is,

[tex]\Delta \theta = \frac{x}{D}[/tex]

Where,

x = The length of the arc

D = Distance

From known data we know that Jupiter's diameter is,

[tex]x_J = 1.43*10^8m[/tex]

[tex]D = 20*9.4608*10^{15}[/tex]

[tex]\lambda = 600*10^{-9}m[/tex]

Replacing we have that,

[tex]\frac{x}{D} = 1.22\frac{\lambda}{d}[/tex]

[tex]\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}[/tex]

Re-arrange to find d,

[tex]d = 968.5m = 0.968Km[/tex]

Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.

Final answer:

The angular acceleration of the disk is calculated using the torque exerted by the applied force and the moment of inertia of a uniform disk. The calculated angular acceleration is 5.33 rad/s².

Explanation:

The question involves calculating the angular acceleration of a rotating disk when a force is applied tangentially at its rim. To find the angular acceleration, we first need to calculate the torque exerted by the force and then use the moment of inertia of the disk to find the angular acceleration.

Given:

The torque (τ) exerted by the force is the product of the force and the radius at which it is applied (torque = force × radius), so:

τ = F × r = 4.0 N × 0.15 m = 0.60 N·m

The moment of inertia (I) for a uniform disk is given by (1/2)mr². Thus:

I = (1/2) × 5.0 kg × (0.15 m)² = 0.1125 kg·m²

Using the formula for angular acceleration (α = torque / moment of inertia), we get:

α = τ / I = 0.60 N·m / 0.1125 kg·m² = 5.33 rad/s²

Therefore, the angular acceleration of the disk is 5.33 rad/s².