A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as heat to the surroundings.) The average wavelength of the visible light is about 600 nm, so make the simplifying assumption that all the light has this wavelength. How many visible-light photons does the bulb emit per second?

The angular momentum of the ice skater spinning at 6.00 rev/s is 15.82 kg · m²/s. When the skater reduces their rate of spin to 1.95 rev/s, the moment of inertia is 1.29 kg · m². If the skater keeps their arms in and allows friction with the ice to slow them to 3.00 rev/s in 23.0 seconds, the average torque exerted is -0.343 N · m (indicating the direction of the torque).

(a) The angular momentum of the ice skater can be calculated by multiplying the angular velocity (in radians per second) by the moment of inertia (in kg · m²). First, we need to convert the angular velocity from rev/s to rad/s. Since 1 rev = 2π rad, the angular velocity in rad/s is 6.00 rev/s * 2π rad/rev = 37.70 rad/s. Now, we can calculate the angular momentum: Angular momentum = angular velocity * moment of inertia = 37.70 rad/s * 0.420 kg · m² = 15.82 kg · m²/s.

(b) To find the new moment of inertia when the angular velocity decreases to 1.95 rev/s, we can rearrange the formula for angular momentum to solve for moment of inertia. Angular momentum = angular velocity * moment of inertia. Rearranging the equation: Moment of inertia = angular momentum / angular velocity = 15.82 kg · m²/s / (1.95 rev/s * 2π rad/rev) = 1.29 kg · m².

(c) To calculate the average torque exerted when the skater slows to 3.00 rev/s, we can use the equation torque = change in angular momentum / time. First, we need to calculate the change in angular momentum by subtracting the initial angular momentum from the final angular momentum. The initial angular momentum is 6.00 rev/s * 2π rad/rev * 0.420 kg · m² = 15.82 kg · m²/s, and the final angular momentum is 3.00 rev/s * 2π rad/rev * 0.420 kg · m² = 7.91 kg · m²/s. The change in angular momentum is 7.91 kg · m²/s - 15.82 kg · m²/s = -7.91 kg · m²/s. Finally, we can calculate the average torque: Torque = change in angular momentum / time = -7.91 kg · m²/s / 23.0 s = -0.343 N · m (negative sign indicates the direction of the torque).

D. the gravitational force increases by a factor of 4

Explanation:

The magnitude of the gravitational force between two objects is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

Here let's call F the initial gravitational force between the two objects. Later, the distance between the objects is halved, so the new distance is:

[tex]r'=\frac{r}{2}[/tex]

Substituting into the equation, we find the new force:

[tex]F'=G\frac{m_1 m_2}{(r/2)^2}=4(G\frac{m_1 m_2}{r^2})=4F[/tex]

Therefore, the gravitational force increases by a factor of 4.

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Final answer:

The gravitational force between two objects doubles if the distance between them is cut in half.

Explanation:

The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that if the distance between the objects is cut in half, the force will increase by a factor of 4.

For example, let's say the original distance is 10 units. If we cut it in half, the new distance would be 5 units. The force would increase by a factor of (10/5)² = 4, meaning the gravitational force would quadruple.

So the correct answer is, A. the gravitational force doubles.

Answer:

Part A:

[tex]n=8.85*10^{28}m^{-3}[/tex]

Part B:

[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

[tex]N-Au=\frac{Density*Avogadro Number}{atomic weight}[/tex]

[tex]Density = 19.32g/cm^3[/tex]

[tex]Avogadro Number=6.02*10^{23} atoms/mol[/tex]

[tex]Atomic weight=196.97g/mol[/tex]

So:

[tex]n=1.5*\frac{Density*Avogadro Number}{atomic weight}[/tex]

[tex]n=1.5*\frac{19.32*6.02*10^{23}}{196.97}[/tex]

[tex]n=8.85*10^{28}m^{-3}[/tex]

Part B:

[tex]Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}[/tex]

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

[tex]Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}[/tex]

[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]

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

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

Final answer:

The new pressure drop p2 will be equal to the original pressure drop p1 at the original mass flow rate.

Explanation:

The relation between the new pressure drop p2 and the original pressure drop p1 at the original mass flow rate can be determined using Poiseuille's equation. Poiseuille's equation states that the pressure drop p2 - p1 is equal to the product of the resistance R and the flow rate Q.

When the length of the pipe is doubled, the resistance R remains constant because the friction factor is constant at high Reynolds numbers. Therefore, the new pressure drop p2 will be equal to the original pressure drop p1.

Answer:

75 W/m²

Explanation:

[tex]I_0[/tex] = Unpolarized light intensity = 800 W/m²

[tex]\theta[/tex] = Angle between filters = 30°

Intensity of light after passing through first polarizer

[tex]I=\frac{I_0}{2}\\\Rightarrow I=\frac{800}{2}\\\Rightarrow I=400\ W/m^2[/tex]

Intensity of light after passing through second polarizer

[tex]I_1=Icos^2\theta\\\Rightarrow I_1=400\times cos^2(30)\\\Rightarrow I_1=300\ W/m^2[/tex]

Intensity of light after passing through third polarizer

[tex]I_2=I_1cos^2\theta\\\Rightarrow I_2=300\times cos^2(90-30)\\\Rightarrow I_1=75\ W/m^2[/tex]

The intensity of the light passing through the stack of polarizing sheets is 75 W/m²

The intensity of the light passing through three polarizing sheets, with the first and the third being perpendicular and the second making an angle of 30 degrees with the first, is zero, due to the rules of light polarization.

Unpolarized light is composed of many rays having random polarization directions. When it passes through a polarizing filter, it decreases its intensity by a factor of 2, thus the intensity after the first polarizer is 800 W/m^2 /2 = 400 W/m^2. According to Malus's Law, the intensity of polarized light after passing through a second polarizing filter is given as I = Io cos² θ, where Io is the incident intensity (from the first polarizer) and θ is the angle between the direction of polarization and the axis of the filter.

The second polarizer is making an angle of 30 degrees with the first polarizer. Therefore, the intensity after the second polarizer is I = 400 W/m^2 * cos²(30) ≈ 346 W/m^2. The third polarizer is perpendicular to the first polarizer (or, equivalently, it makes an angle of 90 degrees with the second polarizer), so the final intensity is I = 346 W/m^2 * cos²(90), which is 0 because the cosine of 90 degrees is zero. So, no light (intensity=0) passes through the stack of these polarizing sheets.

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Answer:

Radius of the outer most dark fringe is 2.65 cm

Solution:

As per the question:

Radius of curvature of the glass, r = 10.8 m

No. of dark fringes, n = 100

Wavelength of light, [tex]\lambda = 652\ nm = 652\times 10^{- 9}\ m[/tex]

Now,

To calculate the radius R of the outermost ring:

Radius of the dark fringe of nth order is given by:

[tex]R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}[/tex]

[tex]R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm[/tex]

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) T = 0.579 s , b)  T = 0.579 s

Explanation:

The great advantage of wave mechanics is that the general equations for different systems are the same, what changes are the physical parameters involved, the equation of motion is

    x = A cos (wt +φ)

Where w is the angular velocity that is this case for being a solid body is

    w = √ (mg d / I)

Where I is the moment of inertia and d the distance to the pivot point

The moment of inertia for a ruler hold one end is

     I = 1/12 M L²

The lost is related to the frequency and is with the angular velocity

     T = 1 / f

    w = 2π f

    w = 2π / T

    T = 2π / w

    T = 2π √ I / mgd

For our case

    d = L

    T = 2π √(1/12 M L²) / M g L)

    T = 2π √(L/(12 g))

a) wood suppose it is one meter long (L = 1m)

     T = 2π √ (1 / (12 9.8))

     T = 0.579 s

b) metal length (L = 1m)

    T = 2pi RA (1 / (12 9.8))

    T = 0.579 s

The period does not depend on mass but on length

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:

v = 300.6 m/s

Explanation:

given,

mass of bullet(m)  = 10 g = 0.01 kg

mass of block of wood (M)= 5 Kg

speed of the block plus bullet after collision(V) = 0.6 m/s

speed of wood(u) = 0 m/s

original speed of the bullet(v) = ?

using conservation of momentum

m v + Mu = (M + m)V

0.01 x v + 5 x 0 = (5 + 0.01) x 0.6

0.01 x v = (5.01) x 0.6

[tex]v = \dfrac{3.006}{0.01}[/tex]

v = 300.6 m/s

the original speed of the bullet before collision is equal to v = 300.6 m/s

The original speed of the bullet can be determined using the law of conservation of momentum. Applying the formula and solving for the original bullet speed, the bullet was initially traveling at 3 m/s.

This problem is an application of the law of conservation of momentum. The principle states that the total momentum before a collision is equal to the total momentum after the collision. In this case, before the collision, only the bullet has momentum as the block is stationary. After the collision, the momentum is shared between the block and the lodged bullet.

To find the original speed of the bullet, we can set up this equation representing the law of conservation of momentum: m1v1 (before collision) = (m1 + m2) v2 (after collision). Here, m1 = mass of the bullet, v1 = original speed of the bullet, m2 = mass of the block, v2 = final speed of the block and bullet together.

Substituting the given values: (10 g)x = (10 g + 5.00 kg)(0.600 m/s). Solving for x (original bullet speed) yields a value of x = 3 m/s.

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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.

Answer: a) 2.33V b) Ip = 0.0035A, Is = 0.179A

Explanation:

The relationship formula between the Voltage, Number of turns and current in a transformer is given as

Vp/Vs = Np/Ns = Is/Ip

Vp is voltage in the primary coil

Vs is voltage in the secondary coil

Ns is number of turns in the secondary coil

Np is number of turns in the primary coil

Is is current in the secondary coil

Ip is current in the primary coil

a) substituting the values in the formula

Vp/Vs = Np/Ns

120/Vs = 500/9.7

Vs = 120×9.7/500

Vs = 2.33V

b) since secondary now has a resistive load of 13 Ω, Using ohm's law to get current in the secondary

V=IR

Is = Vs/R

Is =2.33/13

Is = 0.179A

To get Ip, we will use the relationship Np/Ns = Is/Ip

500/9.7= 0.179/Ip

Ip = 9.7×0.179/500

Ip = 0.0035A

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:

[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]

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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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:

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:

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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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.

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

Explanation:

Moment of Inertia in physics is the measure of the rotational inertia of an object. It is nothing but the opposition that the body offers to having its speed of rotation about an axis changed by application some external torque.

I= MR^2

M= mass of object R= radius of rotation.

Therefore, MOI depends upon

a)total mass

b)shape and density of the object

c)location of the axis of rotation

The moment of inertia of an object depends on its total mass, the distribution of this mass in relation to the rotation axis (shape and density), and the location of the rotation axis. The higher the moment of inertia, the greater the object's resistance to changes in rotational rate. The parallel axis theorem provides a tool to calculate the moment of inertia for different rotation axes.

The moment of inertia of an object depends on certain factors. Specifically, its total mass, the shape and density of the object, and the location of the axis of rotation. The moment of inertia is defined as the measure of the object's resistance to changes in its rotation rate. The moment of inertia varies depending on how mass is distributed in relation to the axis of rotation.

For instance, an object with more mass distributed farther from its axis of rotation will have a higher moment of inertia. This is seen in a hollow cylinder versus a solid cylinder of the same mass and size. The hollow cylinder has a greater moment of inertia because more of its mass is distributed at a greater distance from its axis of rotation.

A principle known as the parallel axis theorem can be used to find the moment of inertia about a new axis of rotation once it is known as a parallel axis. This theorem involves the center of mass and the distance from the initial axis to the parallel axis.

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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:

  [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:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by [tex]\tau =F\times d=45.5\times 0.0245=1.11475N-m[/tex]

So torque on the rocket will be 1.11475 N -m

To solve this exercise it is necessary to apply the concepts related to Magnetic Flow which is defined through the Gaus Law of Magnetism as the measure of the total magnetic field that passes through a given area.

Vectorially it can be defined as

[tex]\Phi = \vec{A}\vec{B}[/tex]

Where,

A = Area

B = Magnetic Field

A scalar mode can also be expressed as

[tex]\Phi = A*B Cos\theta[/tex]

Where,

[tex]\theta[/tex] is the angle between B and A, at this case the direction are perpendicular then

Our values are given as,

[tex]A = 0.19m^2[/tex]

[tex]B_1 = 0.5T[/tex]

[tex]B_2 = 0.8T[/tex]

The magnetic field component that interests us is the perpendicular therefore

[tex]\Phi = 0.19* 0.5cos(90)[/tex]

[tex]\Phi = 0.095Tm^2[/tex]

Therefore the magnetic flux through this coil is 0.095[tex]Tm^2[/tex]

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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The problem presents a collision scenario where principles of momentum conservation and kinetic energy are applied to determine the final velocity of a striking puck and the fraction of kinetic energy lost during the collision.

The question describes a collision scenario in physics involving two pucks, where conservation of momentum and kinetic energy principles are applied. The question can be resolved by dividing it into two sections: (a) Determination of the velocity of the 0.30-kg puck after the collision, and (b) Determination of the fraction of kinetic energy lost in the collision.

For part (a), we use the law of conservation of momentum, which states that the total linear momentum of an isolated system remains constant, regardless of whether the objects within the system are at rest or in motion. To find the final velocity of the 0.30-kg puck, we subtract the momentum vector of the 0.20-kg puck after the collision from the momentum vector before the collision. For part (b), we first find the initial and final total kinetic energy of the system. The fraction of kinetic energy lost is given by the difference between the initial and final kinetic energies, divided by the initial kinetic energy.

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The principle of conservation of momentum and kinetic energy principles are employed to figure out the final velocity of the 0.30-kg puck and the fraction of kinetic energy lost in the collision. Initial and final velocities of the pucks, as well as their masses, are used in the process.

To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision. To find out the velocity of the 0.30-kg puck after the collision (a), the momentum equation (m₁v₁_initial + m₂v₂_initial = m₁v₁_final + m₂v₂_final) is applied. It is known that initial velocities of 0.30-kg and 0.20-kg pucks are 0 and 9.1 m/s respectively; the final velocity of the 0.20-kg puck is 5.5 m/s at 53° to the x-axis.

To calculate the fractional kinetic energy lost in the collision (b), firstly, the kinetic energy before and after the collision is calculated using the equation KE = 1/2mv₂. The fraction of kinetic energy lost is then (KE_initial - KE_final)/KE_initial.

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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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