A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating and drops onto the first flywheel and sticks to it. Both flywheels have radius R. What is the final angular velocity in terms of the initial angular velocity ω0? Treat each flywheel as a disk (I = (1/2) MR2).

Answer:

[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]

Explanation:

Given,

Magnetic field, B = 0.5 T

Electric field, E = 1.2 x 10⁵ V/m

strength of the magnetic field that separates the ions, Bo=0.750 T

Radius, r = 2.32 cm

Relation of charge to mass ratio is given by

[tex]\dfrac{q}{m}=\dfrac{E}{BB_0R}[/tex]

[tex]m=\dfrac{qBB_0R}{E}[/tex]

Substituting all the values

[tex]m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}[/tex]

[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]

Mass of Li ions is equal to [tex]m = 1.16 \times 10^{-26}\ Kg[/tex]

The mass of the Li ions is approximately [tex]\( 1.64 \times 10^{-26} \)[/tex] kg.

To find the mass of the Li ions, we can use the formula for the radius of the circular path of a charged particle moving in a magnetic field:

[tex]\[ r = \frac{m \cdot v}{q \cdot B} \][/tex]

Where:

- r  is the radius of the circular path

- m is the mass of the ion

- v is the velocity of the ion

- q is the charge of the ion

- B is the magnetic field strength

We can also use the formula for the force experienced by a charged particle moving in both electric and magnetic fields:

[tex]\[ F = q \cdot (E + v \times B) \][/tex]

Since the particle moves in a circular path, the electric force F must be equal to the magnetic force [tex]\( q \cdot v \cdot B \)[/tex], where v is the speed of the particle.

We can rearrange the formulas to solve for m:

[tex]\[ m = \frac{q \cdot r \cdot B}{v} \][/tex]

Now, we need to find \( v \). Since the particle passes through both the electric and magnetic fields without deviation, the forces acting on it must be balanced. Therefore:

[tex]\[ F_{electric} = F_{magnetic} \]\[ q \cdot E = q \cdot v \cdot B \][/tex]

Solving for v:

[tex]\[ v = \frac{E}{B} \][/tex]

Now, substituting v into the equation for m :

[tex]\[ m = \frac{q \cdot r \cdot B}{\frac{E}{B}} \]\[ m = \frac{q \cdot r \cdot B^2}{E} \][/tex]

Now we can calculate \( m \) using the given values:

[tex]\[ m = \frac{(1.6 \times 10^{-19} C) \times (0.0232 m) \times (0.75 T)^2}{1.2 \times 10^5 V/m} \]\[ m= 1.64 \times 10^{-26} kg \][/tex]

Answer:

Spiral galaxy

Explanation:

If a galaxy forms from a protogalactic cloud with a lot of angular momentum, we would expect this galaxy to be a spiral galaxy. This is due to the conservation of angular momentum, where greater angular momentum usually results in a flatter, disk-like shape seen in spiral galaxies. Other factors also affect galaxy formation.

In the subject of astrophysics, the angular momentum of the protogalactic cloud has significant implications for the type of galaxy formed. Specifically, if a galaxy forms from a protogalactic cloud with a high angular momentum, we'd expect this galaxy to be a spiral galaxy.

Angular momentum is a property of rotating bodies, determined by the mass, shape, and spin rate of the object. In the context of galaxies, a protogalactic cloud with higher angular momentum is more likely to flatten out into a spinning disk shape, leading to the formation of a spiral galaxy. This is due to the conservation of angular momentum. In contrast, lower angular momentum could lead to the formation of elliptical galaxies, which are less flat and more randomly arranged.

In the realm of galaxy formation and evolution, these are widely accepted ideas, though it's important to note that other factors, like the gas content of the protogalactic cloud and interactions with other galaxies, can also play a major role.

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Answer:

Explanation:

b ) The problem is based on Doppler's effect of sound

f = f₀ x (V - v₀) /( [tex]V+v_s[/tex])

f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away  , [tex]v_s[/tex] is velocity of source going away

778  = 840 x (340 - 14)/ (340 + [tex]v_s[/tex])

340 + [tex]v_s[/tex] = 341.18

[tex]v_s[/tex] = 1.18 m /s

it will go  away from   the observer or the cyclist.

speed of train = 1.18 m /s

a )

For a stationary observer v₀ = 0

f = f₀ x V  /( [tex]V+v_s[/tex])

= 840 x 340 / (340 + 1.180)

= 837 Hz

Final answer:

The stationary observer between the train and the bicycle hears a frequency of 809 Hz, and the train is moving away from the bicycle at 42 m/s.

Explanation:

A stationary observer located between the train and the bicycle would hear a frequency of 809 Hz as the train approaches and passes by.

The speed of the train can be calculated to be 42 m/s moving away from the bicycle.

Answer:

B(2 pi r) = uo i = uo (J pi r^2)

Magnetic field, B = μo Jr/2

Explanation:

The magnitude of the magnetic field inside the wire at a distance r1 < R inside the wire is;

B = ½μ_o × Jr

We want to find the magnitude of the magnetic field inside the wire at a distance r1 < R.

The formula for Magnetic field at a distance r located inside a conductor of radius R is given by Ampere's circuital law as;

B × 2πr = μ_o × Jπr²

Where;

B is magnitude of magnetic field

μ_o is a constant known as magnetic permeability

J is current density

Thus;

B × 2πr = μ_o × Jπr²

πr will cancel out from both sides to give;

2B = μ_o × Jr

B = ½μ_o × Jr

Thus, in conclusion them magnitude of the magnetic field is ½μ_o × Jr

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Answer:

1.0416 m∧3/sec

Explanation:

check the pictures below for the solution

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

[tex]p_f=p_i\\mv_1+Mv_2=(m+M)v[/tex]

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

[tex](0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}[/tex]

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

[tex]E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2[/tex]

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

[tex](0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m[/tex]

hence, the heigth is 68cm

Answer:

64.57

Explanation:

We are given that

For decreasing measured sound intensity level=36.2 dB

We have to find the  factor by which the distance increase.

Let initial distance=x

Final distance=x'

According to question

[tex]36.2=10log(\frac{x'^2}{x^2})[/tex]

[tex]36.2=10log(\frac{x'}{x})^2[/tex]

[tex]36.2=10\times 2log\frac{x'}{x}[/tex]

[tex]log\frac{x'}{x}=\frac{36.2}{20}=1.81[/tex]

[tex]\frac{x'}{x}=10^{1.81}[/tex]

[tex]x'=10^{1.81}x=64.57x[/tex]

Hence, the distance is increases by factor of 64.57

Answer:

0.16 s

Explanation:

• Falling from rest (V_block= 0 m/s) the block attains a final velocity V_block before colliding with the bullet. This velocity is given by Equation 2.4 as

V_block(final velocity of block just before hitting) =V_0,block +at

where a is the acceleration due to gravity (a = —9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is

V_block = at

• During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that

(M_bullet+M_block)V_f = M_bullet*V_bullet+ M_block*V_block  

Total linear momentum after collision = Total linear momentum  before collision

Here V_f is the final velocity of the bullet/block system after the collision, and V_bullet and V_block  are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that V_f, the final velocity of the bullet/block system after the collision must have the same magnitude as V_block, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that of V_f =-V_block, Substituting this relation and Equation (1) into Equation (2) gives

(M_bullet + M_block)(-at) = M_bullet*V_bullet + M_block(at)

t = -M_bullet*V_bullet/a(M_bullet +2M_block)

 =-(0.0140-kg)*555 m/s/-9.8(0.0140-kg+2(2.42 kg)

 =0.16 s

Final answer:

The student's physics question involves applying the conservation of momentum and kinematic equations to determine the time the block was falling before being struck by a bullet. By considering the bullet's speed, the mass of both the bullet and block, and the motions involved, one can calculate the time of fall.

Explanation:

The student's question involves a physics problem related to conservation of momentum and projectile motion. The bullet's speed when it strikes the block, and the subsequent motion of the block and bullet system provide key information. To solve for the time t that the block was falling, we need to use the principles of physics that dictate how objects move under the influence of gravity and how they interact in collisions.

To find the time t, we'll take the following steps:

Use the conservation of momentum to find the velocity of the block and bullet immediately after the collision.

Use kinematic equations to relate this velocity to the maximum height reached by the block and bullet, taking into account the direction reversal.

Use the kinematic equations again to find the time t during which the block was falling before the collision.

Assuming the collision is perfectly inelastic, the bullet embeds itself in the block, so we have a combined mass moving upward after the collision. This combined mass will move up to the original height due to the conservation of energy principle, as the initial kinetic energy is converted entirely to gravitational potential energy at the peak of the ascent after which it momentarily comes to a halt.

Given that,

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules. It is moving upward with a kinetic energy of 50 Joules.

We need to find the maximum height h reached by the ball. Let at a height of 10 meters, it has a potential energy of 50 Joules. So,

[tex]mgH=50\\\\mg=\dfrac{50}{h}\\\\mg=\dfrac{50}{10}\\\\mg=5\ N[/tex] ........(1)

Let at a height of h m, it reaches to a maximum height. at this point, it has a total of 100 J of energy. So,

[tex]mgh=50+50\\\\mgh=100\\\\h=\dfrac{100}{5}\\\\h=20\ m[/tex]

So, the correct option is (E) "h = 20 m".

The maximum height reached by the ball is about 20 meters, as determined by using the conservation of mechanical energy principle, considering that the total mechanical energy at the height of 10 meters was 100 Joules.

To solve for the maximum height reached by the ball, we can use the conservation of mechanical energy principle, which states that the total mechanical energy (potential energy + kinetic energy) of the ball remains constant in the absence of air friction.

At 10 meters above the ground, the ball has a potential energy (PE) of 50 Joules and a kinetic energy (KE) of 50 Joules. Therefore, the total mechanical energy at that height is:

PE + KE = 50 J + 50 J = 100 J

As the ball rises, its kinetic energy is converted into potential energy until the kinetic energy becomes zero at the maximum height. The total mechanical energy at maximum height will be equal to the potential energy:

PE at maximum height = total mechanical energy = 100 J

Using the formula for gravitational potential energy, PE = mgh (where m is mass, g is the acceleration due to gravity (9.81 m/s²), and h is the height), and knowing that the PE at 10 meters is 50 J, we can find the mass of the ball:

50 J = m * 9.81 m/s² * 10 m

m = 50 J / (9.81 m/s² * 10 m) = 0.51 kg

With the mass of the ball, we can now calculate the maximum height using the total mechanical energy:

100 J = 0.51 kg * 9.81 m/s² * h

h = 100 J / (0.51 kg * 9.81 m/s²) ≈ 20 meters

Therefore, the maximum height h reached by the ball is about 20 meters.

B. The Earth radiates an amount of energy into space equal to the amount it receives.

Part of the solar energy is reflected by the Earth into space, this is known as albedo. The other part of the energy radiated by the Earth in the form of infrared radiation, is absorbed by the greenhouse gases, which cause most of this infrared radiation to be emitted into space. Therefore, the net flow of energy is zero.

Answer:

The answer is 0.001 m

Explanation:

Solution

Recall that,

A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at = 1200°C

The diffusion coefficient for nitrogen in steel at this temperature is =6 * 10-11 m2/s

The diffusion flux is = 1.2 *10^-7 kg/m2s

Concentration of nitrogen in the steel at the high-pressure surface is= 4 kg/m3.

The high-pressure side will the concentration is estimated to be = 2.0 kg/m3

Now,

The flux = -D dC/dx

1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx

∫ˣ₀ dx = -5x^10-4 ∫²₄ dC

so,

x = (2-4) kg/m3 (-5x10-4 m4/kg)

where x = .001 m

Therefore x = 0.001 m

The distance from this high-pressure side will the concentration is  0.001 m

Since the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, the diffusion flux is found to be 1.2x 10-7 kg/m2 -s.

We know that

The flux =[tex]-D\ dC\div dx[/tex]

So,

1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx

dC/dx = -2000

Now

-2000 = 4 - 2/0-x_B

x_B = 2/2000

x_B = 1*10^-3m

x_B = 0.001m

Hence, The distance from this high-pressure side will the concentration is  0.001 m

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Answer:

The answer is

D. 1230j

Explanation:

When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force

given data

Mass of gun M=10kg

Mass of bullet m=20g----kg=20/1000 =0.02kg

Propulsive speed of bullet = 350m/s

Hence the moment of the bullet will be equal and opposite to that of the gun

mv=MV

where V is the recoil velocity which we are solving for

V=mv/M

V=0.02*350/10

V=7/10

V=0.7m/s

The energy contained in the bullet can be gotten using

KE=1/2m(v-V)²

KE=1/2*0.02(350-0.7)²

KE=1/2*0.02(349.3)²

KE=1/2*0.02*122010.49

KE=1/2*2440.20

KE=1220.1J

roughly the energy is 1230J

Answer:

Option B and Option D are true

Explanation:

We are given;

Number of atoms in block A = 800

Energy content in block A = 20 quanta

Number of atoms in block B = 200

Energy content in block B = 80 quanta

The energy of a system which is an extensive quantity,depends on the mass or number of moles of the system. However, at equilibrium, the energy density of the two copper blocks will be equal. That is, each atom of Cu in the two blocks will, on average, have the same energy. Because block A has 4 times more atoms than block B, it will have 4 times more quanta of energy. Thus, option B is therefore true while option A is false.

Temperature is a measure of the average kinetic energy of the atoms in a material. Now, if each atom in blocks A and B have the same average energy, then the temperatures of blocks A and B will be equal at equilibrium. Thus, option D is true.

Entropy of a system is an extensive quantity that depends on the the mass or number of atoms in the system. Because block A is bigger than block B, it will have higher entropy. However, that the specific entropy (the entropy per mole or per unit mass) is an intensive quantity -- it is independent of the size of a system. The molar entropy of blocks A and B are equal at equilibrium. Thus option C is false.

For the given condition in the paragraph, the statements (a) and (c) are false. And the statements given in option (b) and (d) are true.

Given data:

Number of atoms in Block A is, 800 atoms.

Initial amount of energy in Block A is, 20 quanta.

Number of atoms in Block B is, 200 atoms.

Initial amount of energy in Block B is, 80 quanta.

We will use the concept of Energy to relate it with the temperature, to solve the given problem. Also, there is some glance about the equilibrium condition as discussed:

Thus, we conclude that for the given condition, the statements (a) and (c) are false. And the statements given in option (b) and (d) are true.

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Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a  , a is separation of slits

Puting the values given

.51 x 10⁻³ =  502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

[tex]I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2[/tex]

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

[tex]I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2[/tex]

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

[tex]I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2[/tex]

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

[tex]I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}[/tex]

[tex]\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}[/tex]

[tex]\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s[/tex]

4)Kinetic energy before:

[tex]E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J[/tex]

Kinetic energy after:

[tex]E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J[/tex]

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) [tex]a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2[/tex]

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

Solution:

a) We know acceleration due to gravity, g = GM/r²

Differential change, dg/dr = -2GM/r³

Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²

My height, h=dr = 1.7 m

Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²

or,   dg/dr = -10/1.7 = -2GM/(100GM/c²)³

or,     5.9*100³*G²*M² = 2c⁶

or,   M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms

Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.

b) This limit an upper limit ,we can tolerate smaller masses only.

Answer:

39 m

Explanation:

We are given that

Wavelength=[tex]\lambda=400 nm=400\times 10^{-9} m[/tex]

1 nm=[tex]10^{-9} m[/tex]

y=1 km=[tex]1000 m[/tex]

1 km=1000 m

Earth mars distance =x=80 million Km=[tex]80\times 10^9 m[/tex]

1million km=[tex]10^9 m[/tex]

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

[tex]sin\theta=\frac{y}{x}[/tex]

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

[tex]\frac{1000}{80\times 10^9}=\frac{1.22\times 400\times 10^{-9}}{d}[/tex]

[tex]d=\frac{1.22\times 400\times 10^{-9}\times 80\times 10^9}{1000}[/tex]

[tex]d=39.04 m\approx 39 m[/tex]

Diameter of mirror =39 m

To achieve a 1.0-km resolution on Mars with light of a 400 nm wavelength, the size of the space telescope's mirror would need to be approximately 6 mm. This calculation is based on diffraction limit theory and may not be precisely accurate in practical application due to other potential limiting factors.

This question is related to the resolution of a telescope which is primarily influenced by the aperture size and wavelength of light being observed. The bigger the telescope aperture (the diameter of the primary mirror), the greater the resolution as more light can be collected. The formula for the resolution limit due to diffraction (the smallest distinguishable detail) can be given by R = 1.22λ/D where λ is the wavelength of light and D is the diameter of the telescope's main mirror.

Here the wavelength given is 400 nm or 400 x 10^-9 m, and we need to calculate the mirror diameter required for a resolution of 1.0-km on Mars from Earth. Let's convert the resolution limit from kilometers to meters (which gives us 1000m), and then to radians (using the Earth-Mars distance), which results in an angle of 1.25x10^-11 radians approximately.

Substituting these values in the formula, we can solve for D and find that the telescope mirror size needed would be approximately 0.006 m or 6 mm for this resolution. Do note, that this is a theoretical value, in reality, the size might need to be larger due to factors like the diffraction limit, non-uniformities in mirrors, or aberrations in lenses.

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Answer:

The gravity at planet X is  [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]

The value of [tex]\alpha[/tex] on the moon is  [tex]\alpha = 730.38 \ revolutions[/tex]

Explanation:

  From the question we are told that

        The clocks hour hand makes [tex]\alpha[/tex] revolution every 1 hour which is 3600 sec

this implies that the time peroid for 1 revolution would be  [tex]= \frac{3600}{\alpha }sec[/tex]

   The peroid  for a pendulum is mathematically represented as

            [tex]T = 2 \pi\sqrt{\frac{L}{g} }[/tex]

Where L is the pendulum length

            g is the acceleration due to gravity

Let assume that we have a pendulum that counts in second on earth

   This implies that its peroid would be = 2 second

   i.e one second to swimg forward and one second to swing back to its original position

  Now the length of this pendulum on earth is

          [tex]L = \frac{gT^2}{4 \pi^2}[/tex]         [Making L the subject in above equation]

 Substituting values

        [tex]L = \frac{9.8 * (2)^2}{4 * (3.142)^2}[/tex]

           [tex]= 1[/tex]

When the same pendulum is taken to planet X  the peroid would be

                 [tex]T = \frac{3600}{\alpha }[/tex]

Recall this value was obtained above for 1 revolution (from start point to end point back to start point)

      So the acceleration due to gravity on this planet would be mathematically represented as

                [tex]g = \frac{4 \pi L }{T^2}[/tex]   [making g the subject in the above equation]

     substituting values

               [tex]g = \frac{4 * 3.142^2 * 1}{[ \frac{3600}{\alpha } ]^2}[/tex]

                    [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]

On moon the acceleration due to gravity has a constant value of

     [tex]g = 1.625 m/s^2[/tex]

The period of this pendulum  on the moon can be mathematically evaluated as

      [tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]

 substituting value

     [tex]T = 2 *3.142 \sqrt{\frac{1}{1.625} }[/tex]

        [tex]= 4.929s[/tex]

 given that  

       [tex]1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s[/tex]  {Note 1 revolution takes a peroid }

Making [tex]\alpha[/tex] the subject of the formula

           [tex]\alpha =\frac{3600}{4.929}[/tex]

              [tex]\alpha = 730.38 \ revolutions[/tex]

Answer:

The circuit breaker will trip because the current drawn is in excess of its capacity

Explanation:

The 20 A circuit breaker is designed to not carry current in excess of its capacity.

Total power demand on power strip = 900 + 990 + 650 = 2540 w

Let us assume a standard American voltage outlet of 120 V

Recall that electric power is

P = I x V

Where I is current, and

V = voltage.

For the 2540 W power drawn, the current I is

I = P/V = 2540/120 = 21.16 A

This is 1.16 A in excess of its capacity.

Answer:

2.5kN.m

Explanation:

Torque is directly proportional to pitch diameter

= Ta/Tb= Da/Db

=120/Tb= 0.25/0.5

Tb= 2.469kN.m approx 2.5kN.m

Answer:

Explanation:

answer explain below

The ventriloquist uses the perceptual principle known as the McGurk Effect to create the illusion that words are coming from their dummy's mouth. This involves syncing their speech with the movements of the dummy's mouth and mastering aspects of speech production. We perceive the words to be coming from the dummy due to the combined audiovisual cues.

A ventriloquist convinces us that words are coming from their dummy's mouth by making use of the perceptual principle known as the McGurk Effect. This audiovisual speech perception demonstrates that we don't just hear speech but also 'see' it. The ventriloquist creates the illusion by carefully syncing the opening and closing of the dummy's mouth with their own speech. This overlaid speech and vision information persuades us to perceive the sound as coming from the dummy.

Human speech itself is produced by shaping the cavity formed by the throat and mouth, the vibration of vocal cords, and using the tongue to adjust the fundamental frequency and combination of those sounds. Thus, excellent ventriloquists master these aspects of speech production without visibly moving their own mouth or lips, reinforcing the illusion.

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Answer:

[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex]

[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex]

Explanation:

The electric field equation of a electromagnetic wave is given by:

[tex]E=E_{max}(kx-\omega t)[/tex] (1)

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

[tex]k=\frac{2\pi}{\lambda}[/tex]            

[tex]k=8.98*10^{6} [rad/m][/tex]      

And the relation between λ and f is:                

[tex]c=\lambda f[/tex]

[tex]f=\frac{c}{\lambda}[/tex]

[tex]f=\frac{3*10^{8}}{700*10^{-9}}[/tex]

[tex]f=4.28*10^{14}[/tex]

The angular frequency equation is:

[tex]\omega=2\pi f[/tex]

[tex]\omega=2\pi*4.28*10^{14}[/tex]

[tex]\omega=2.69*10^{15} [rad/s][/tex]

Therefore, the E equation, suing (1), will be:

[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex] (2)

For the magnetic field we have the next equation:

[tex]B=B_{max}(kx-\omega t)[/tex] (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

[tex]E_{max}=cB_{max}[/tex]

[tex]B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}[/tex]

[tex]B_{max}=1.17*10^{-8}T[/tex]

Putting this in (3), finally we will have:

[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex] (4)

I hope it helps you!

Final answer:

To find the equations of the fields for an electromagnetic wave in the red spectrum with given parameters, calculate the frequency using the speed of light and wavelength, then apply it to the sinusoidal equations representing the electric and magnetic fields.

Explanation:

To determine the equations for the electric and magnetic fields (E and B, respectively) of an electromagnetic wave in the visible red spectrum with a wavelength (λ) of 700 nm and a peak electric field magnitude (Ē) of 3.5 V/m, we first need to calculate the frequency (f) of the wave.

Since the speed of light (c) is approximately 3 × 108 m/s, the frequency can be calculated by using the relationship c = f × λ. After finding the frequency, we can then write down the equation for the electric field E(x, t). Assuming that the wave is propagating in the +x direction and that the electric field oscillates in the y-direction, with no initial phase change, the equation for the electric field is:

E(x, t) = Ē × sin(2π(ft - x/λ))

To find the associated magnetic field B, we use the fact that the magnitudes of E and B are related by the speed of light c, such that B = E/c. Since electromagnetic waves have their electric and magnetic fields perpendicular to each other, if E is in the y-direction, B will be in the z-direction. The equation for the magnetic field is:

B(x, t) = B × sin(2π(ft - x/λ))

Where B is the peak magnetic field strength.

Answer:

[tex]T_{1}=105.38 N[/tex]

[tex]T_{2}=187.34 N[/tex]

Explanation:

Applying the second Newton's law for the first box we have.

[tex]-f_{1f}+T_{1}=m_{A}a[/tex]

[tex]-\mu_{k}N_{1}+T_{1}=m_{A}a[/tex]

We know that the normal force is the product between the weight and the kinetic friction, so we have:

[tex]-\mu_{k}m_{A}g+T_{1}=m_{A}a[/tex]

Now we can find T₁:

[tex]\mu_{k}m_{A}g+m_{A}a=T_{1}[/tex]

The acceleration is the same for both boxes.

[tex]T_{1}=m_{1}(\mu_{k}g+a)[/tex]

[tex]T_{1}=18*(0.240*9.81+3.5)[/tex]

[tex]T_{1}=105.38 N[/tex]

Now let's analyze the forces of the second box.

[tex]-f_{2f}-T_{1}+T_{2}=m_{B}a[/tex]

[tex]-\mu_{k}m_{B}g-T_{1}+T_{2}=m_{B}a[/tex]

Let's solve it for T₂.

[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]

[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]

[tex]T_{2}=m_{B}(a+\mu_{k}g)+T_{1}[/tex]

[tex]T_{2}=14(3.5+0.240*9.81)+105.38[/tex]

[tex]T_{2}=187.34 N[/tex]

I hope it helps you!

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

Answer:

Explanation:

Given that, .

Frequency

f = 60Hz

Number of turns

N = 125turns

Surface area of coil

A = 3 × 10^-2 m²

Magnetic field

B = 0.12T

Voltage peak to peak? I.e the EMF

EMF is given as

ε = —dΦ/dt

Where Φ is magnetic flux and it is given as

Φ = NBA Cosθ

Where N is number of turns

B is magnetic field

A is the cross sectional area

And θ is the resulting angle from the dot product of area and magnetic field

Where θ =ωt and ω = 2πf

Then, θ = 2πft

So, your magnetic flux becomes

Φ = NBA Cos(2πft)

Now, dΦ / dt = —NBA•2πf Sin(2πft)

dΦ / dt = —2πf • NBA Sin(2πft)

So, ε = —dΦ/dt

Then,

ε = 2πf • NBA Sin(2πft)

So, the maximum peak to peak emf will occur when the sine function is 1

I.e Sin(2πft) = 1

So, the required peak to peak emf is

ε = 2πf • NBA

Substituting all the given parameters

ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2

ε = 169.65 Volts

The peak to peak voltage is 169.65 V

The required value of peak output voltage is 169.65 Volts.

Given data:

The frequency of ac generator is, f = 60 Hz.

The number of turns of coil is, n = 125 turns.

The area of each coil is, [tex]A =3.0 \times 10^{-2} \;\rm m^{2}[/tex].

The strength of magnetic field is, B = 0.12 T.

To start with this problem, we need to find the peak emf first. The peak output voltage is nothing but the value of this peak emf only. The expression for the  peak EMF is given as,

ε = —dΦ/dt

here,

Φ is magnetic flux and it is given as

Φ = NBA Cosθ

Here,

θ is the resulting angle from the dot product of area and magnetic field

Where θ =ωt and ω = 2πf

Then, θ = 2πft

So, the expression for the magnetic flux becomes,

Φ = NBA Cos(2πft)

Now, dΦ / dt = —NBA•2πf Sin(2πft)

dΦ / dt = —2πf • NBA Sin(2πft)

So, ε = —dΦ/dt

Then,

ε = 2πf • NBA Sin(2πft)

So, the maximum peak to peak emf will occur when the sine function is 1

 Sin(2πft) = 1

So, the required peak to peak emf is

ε = 2πf • NBA

Substituting all the given parameters

ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2

ε = 169.65 Volts

Thus, we can conclude that the required value of peak output voltage is 169.65 Volts.

Learn more about the peak voltage here:

https://brainly.com/question/14789569

Answer:

The wavelength is  [tex]\lambda = 706nm[/tex]

Explanation:

From the question we are told that

    The refractive index of the glass is [tex]n__{glass}} = 1.52[/tex]

    The thickness of film is [tex]D = 127nm = 127*10^{-9}m[/tex]

     The refractive index of film [tex]n__{film}} = 1.39[/tex]

     The refractive index of air is [tex]n__{air}} = 1.00[/tex]

Generally the thickness of the film can be obtained mathematically from this expression

               [tex]D = \frac{\lambda}{4 * n__{film}} }[/tex]

Where [tex]\lambda[/tex] is the wavelength

           Making the wavelength the subject of the formula

                      [tex]\lambda = 4 * n__{film}} * D[/tex]

Substituting values

                     [tex]\lambda = 4 *1.39 * 127 *10^{-9}[/tex]

                       [tex]\lambda =7.06 *10^{-7}m = 706nm[/tex]

Answer:

[tex]\lambda = 706.12 nm[/tex]

Explanation:

The optical path length for the reflection of light = [tex]2 n_{film} t[/tex]

For destructive interference,  [tex]2 n_{film} t = \frac{\lambda}{2}[/tex]

The thickness of the anti-reflective film = 127 nm

The largest wavelength of the reflected light, [tex]\lambda = 4n_{film} t[/tex]

[tex]\lambda = 4 * 1.39 *127 * 10^{-9}[/tex]

[tex]\lambda = 706.12 * 10^{-9} m\\\lambda = 706.12 nm[/tex]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  [tex]v_A= 4m/s[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]L = 1m[/tex]

     The initial speed of block A is [tex]u_A[/tex]

     The final speed of block A is  [tex]v_A = \frac{1}{2}u_A[/tex]

      The initial speed of block B is [tex]u_B = 0[/tex]

      The mass of block A  is  [tex]m_A = 7kg[/tex]  gh

      The mass of block B is  [tex]m_B = 2 kg[/tex]

According to the principle of conservation of momentum

       [tex]m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}[/tex]

Since block B at initial is at rest

       [tex]m_A u_A = m_Bv_B + m_A \frac{u_A}{2}[/tex]

      [tex]m_A u_A - m_A \frac{u_A}{2} = m_Bv_B[/tex]

          [tex]m_A \frac{u_A}{2} = m_Bv_B[/tex]

  making [tex]v_B[/tex] the subject of the formula

             [tex]v_B =m_A \frac{u_A}{2 m_B}[/tex]

Substituting values

               [tex]v_B =\frac{7 u_A}{4}[/tex]  

This [tex]v__B[/tex] is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of [tex]v__B'[/tex] at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   [tex]a= \frac{v^2_{B}'}{L}[/tex]

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         [tex]T + mg = m \frac{v^2_{B}'}{L}[/tex]

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 [tex]\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L[/tex]

From above  

                [tex](T + mg) L = m v^2_{B}'[/tex]

Substitute this into above equation

             [tex]\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L[/tex]  

             [tex]\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L[/tex]

          [tex]\frac{49 mv_A^2}{16} = T + 5mgL[/tex]

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      [tex]\frac{49 mv_A^2}{16} = 5mgL[/tex]

making  [tex]v_A[/tex] the subject

            [tex]v_A = \sqrt{\frac{80mgL}{49m} }[/tex]

substituting values

          [tex]v_A = \sqrt{\frac{80* 9.8 *1}{49} }[/tex]

              [tex]v_A= 4m/s[/tex]

Answer

D, 2t

Explanation:

See attached file

"2t" will be the path length difference between Wave 1 and Wave 2. A complete solution is below.

According to the question,

The path difference between Wave 1 and 2 will be:

When [tex]\Theta = 0^{\circ}[/tex],

→ [tex]P.d= \Delta x[/tex]

         [tex]= AB+BC[/tex]

hence,

Path difference,

= [tex]t+t[/tex]

= [tex]2t[/tex]

Thus the above answer i.e., "option d" is appropriate.

Learn more:

https://brainly.com/question/13958079

Answer:

hii bro.... from India. ....i am from Bilaspur cg