rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 seconds. Determine the distance Rickey slides across the ground before touching the base.

Answer:1.6 rad/s

Explanation:

Given

moment of Inertia  of disk [tex]I=5 kg-m^2[/tex]

radius of disc [tex]r=0.25 m[/tex]

Force [tex]F=8 N[/tex]

Torque [tex]T=I\alpha =F\cdot r[/tex]

[tex]5\times \alpha =8\times 0.25[/tex]

[tex]\alpha =0.4 rad/s^2[/tex]

using

[tex]\theta =\omega _0\times t+\frac{\alpha t^2}{2}[/tex]

[tex]\pi =0+\frac{0.4t^2}{2}[/tex]

[tex]2\pi =0.4t^2[/tex]

[tex]t^2=5\pi [/tex]

[tex]t=\sqrt{5\pi }[/tex]

[tex]t=3.96 s[/tex]

[tex]\omega =\omega _0+\alpha t[/tex]

[tex]\omega =0+0.4\times 3.96[/tex]

[tex]\omega =1.58 rad/s\approx 1.6 rad/s[/tex]

The correct answer is (3) 1.6 rad/s. The angular velocity of the disk after it has turned through half a revolution is approximately [tex]\(1.6 \, \text{rad/s}\)[/tex] .

To solve this problem, we can use the relationship between torque, rotational inertia, and angular acceleration, which is given by the equation:

[tex]\[\tau = I \alpha\][/tex]

 where \(\tau\) is the torque, \(I\) is the rotational inertia, and \(\alpha\) is the angular acceleration.

 First, we calculate the torque \(\tau\) caused by the force \(F\) applied tangentially to the rim of the disk at a distance \(r\) from the axis of rotation:

[tex]\[\tau = F \times r\][/tex]

Given that [tex]\(F = 8.0 \, \text{N}\) and \(r = 0.25 \, \text{m}\),[/tex]we have:

[tex]\[\tau = 8.0 \, \text{N} \times 0.25 \, \text{m} = 2.0 \, \text{Nm}\][/tex]

Now, we know the rotational inertia \(I\) of the disk is [tex]\(5.0 \, \text{kg} \cdot \text{m}^2\)[/tex]. Using the torque and rotational inertia, we can find the angular acceleration \(\alpha\):

[tex]\[2.0 \, \text{Nm} = 5.0 \, \text{kg} \cdot \text{m}^2 \times \alpha\][/tex]

[tex]\[\alpha = \frac{2.0 \, \text{Nm}}{5.0 \, \text{kg} \cdot \text{m}^2} = 0.4 \, \text{rad/s}^2\][/tex]

 Next, we use the equation for angular displacement \(\theta\) when an object starts from rest and has a constant angular acceleration:

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

We are given that the disk has turned through half a revolution, which is \(\pi\) radians (since one full revolution is \(2\pi\) radians). We can now solve for the time \(t\) it takes to achieve this angular displacement:

[tex]\[\pi = \frac{1}{2} \times 0.4 \, \text{rad/s}^2 \times t^2\[/tex]

[tex]\[t^2 = \frac{\pi}{0.2 \, \text{rad/s}^2}\][/tex]

[tex]\[t = \sqrt{\frac{\pi}{0.2 \, \text{rad/s}^2}}\][/tex]

[tex]\[t \approx \sqrt{\frac{3.14159}{0.2}} \approx \sqrt{15.70795} \approx 3.96 \, \text{s}\][/tex]

 Finally, we can find the angular velocity \(\omega\) after this time using the equation:

[tex]\[\omega = \alpha t\][/tex]

[tex]\[\omega = 0.4 \, \text{rad/s}^2 \times 3.96 \, \text{s}\][/tex]

[tex]\[\omega \approx 1.584 \, \text{rad/s}\][/tex]

 Rounding to two significant figures, we get:

[tex]\[\omega \approx 1.6 \, \text{rad/s}\][/tex]

 Therefore, the angular velocity of the disk after it has turned through half a revolution is approximately[tex]\(1.6 \, \text{rad/s}\)[/tex], which corresponds to option (3).[tex]\[\omega = \alpha t\][/tex]

Answer:

292.3254055 W/m²

469.26267 V/m

[tex]1.56421\times 10^{-6}\ T[/tex]

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = [tex]\frac{d}{2}=\frac{7}{2}=3.5\ cm[/tex]

[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]

The intensity is given by

[tex]I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2[/tex]

5% of this energy goes to the visible light so the intensity is

[tex]I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2[/tex]

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

[tex]u=\frac{1}{2}\epsilon_0E^2[/tex]

Energy density is also given by

[tex]\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m[/tex]

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

[tex]B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T[/tex]

The amplitude of the magnetic field at this surface is [tex]1.56421\times 10^{-6}\ T[/tex]

The intensity of visible light at the surface of the bulb is approximately 920 W/m². The amplitude of the electric field at this surface, for a sinusoidal wave with this intensity, is approximately 600 V/m. The amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity, is approximately 2 x 10^-6 T.

A 90-W incandescent light bulb radiates only about 5% of the energy as visible light, this gives 0.05 * 90 = 4.5 W as the power of the visible light. The intensity in watts per square meter is given by power divided by surface area. The surface area of a sphere is 4πR², where R represents the radius of the sphere (3.5 cm = 0.035 m in this case). Thus, the visible light intensity I is given by: I = 4.5W / (4π*0.035²) ≈ 920 W/m².

For a sinusoidal wave, the amplitude E0 of the electric field is related to the intensity I by E0 = sqrt (2µ0cI), where µ0 is the permittivity of free space (8.85x10^-12) and c is the speed of light (3x10^8 m/s). Using the calculated intensity, we find E0 ≈ 600 V/m.

The amplitude B0 of the magnetic field at the light bulb surface related to the electric field's amplitude through the relation B0 = E0 / c, so B0 ≈ 600V/m / 3x10^8 m/s ≈ 2x10^-6 T (Tesla).

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Answer:

(A) 4.09 kg/s

(B) 589.9 m/s

(C)   0.0008707 m^{3} =  8.71 cm^{2}

Explanation:

inlet pressure of steam (P1) = 4 MPa

inlet temperature of steam (T1) = 400 degree celcius

inlet velocity (V1) = 60 m/s

outlet pressure (P2) = 2 MPa

outlet temperature (T2) = 300 degree celcius

inlet area (A1) = 50 cm^{2} = 0.005 m^{2}

rate of heat loss (Q) = 75 kJ/s

(A) mass flow rate (m) = \frac{A1 x V1}{α1}

where the initial specific volume (α1) for the given temperature and pressure is gotten from tables A-6 = 0.07343 m^3/kg

m = \frac{0.005 x 60}{0.07343}

m = 4.09 kg/s

(B) we can get the outlet velocity using the energy balance equation

  E in = E out

   m(h1 + \frac{(V1)^{2}}{2}) =  m(h2 + \frac{(V2)^{2}}{2})

V2 = [tex]\sqrt{2(h1 - h2) +(V1)^{2} - 2\frac{Q}{m}[/tex]

where h1 and h2 are the enthalpies and are gotten from table A-6

V2 = [tex]\sqrt{2 x 1000 x(3214.5 - 3024.2) +(60)^{2} - 2\frac{75 x 1000}{4.09}[/tex]

V2 = 589.9 m/s  

(C) the outlet area is gotten from mass flow rate (m) = \frac{A2 x V2}{α}

  A2 = (α2 x m) / V2

where the initial specific volume (α2) for the given temperature and pressure is gotten from tables A-6 = 0.12552 m^3/kg

A2 = (0.12552 x 4.09) / 589.5 = 0.0008707 m^{3} =  8.71 cm^{2}

To solve this problem on thermodynamics and fluid dynamics, apply the principles of conservation of mass and energy. First, calculate the mass flow rate using specific volume and continuity equation. Then, considering kinetic and internal energy, calculate the exit velocity. Finally, using mass flow rate and exit velocity, determine the exit area of the nozzle.

The subject of this question falls within the field of Engineering, particularly it relates to thermodynamics and fluid dynamics.

Starting with the given situation, we can see that we have the following known quantities: Initial pressure (P1) = 4 MPa, initial temperature (T1) = 400°C, initial velocity (V1) = 60 m/s, final pressure (P2) = 2 MPa, final temperature (T2) = 300°C, and heat loss Q = 75 kJ/s.

To answer part (a), you would first calculate the specific volume at the given initial state using a steam table, then apply the equation of continuity (ρ1A1V1=ρ2A2V2) to find the mass flow rate.

For part (b), you would follow a similar method but this time utilizing the equation of energy conservation considering both kinetic energy and internal energy of the steam, since the mechanical work is zero and there is heat loss. Considering the energy balance, you can find the exit velocity of the steam.

For part (c), you would use the mass flow rate you found in part (a) and the exit velocity from part (b) in the continuity equation to find the exit area of the nozzle.

Keep in mind that these calculations involve the use of particular thermodynamic values, which might be provided in course materials or general engineering tables.

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Answer:

Orbital

Explanation:

The orbitals is the area or region outside the nucleus where an electron can most likely be found

Answer:

s sublevel

Explanation:

IS a orbital

The time Δt it will take for the bike wheel to come to a complete stop is approximately 23.6 seconds.  

Let's break down the problem step by step:

1. Calculate the angular deceleration (α):

  - We're given the initial angular speed (ω0) as 9.0 rotations per second and the time it takes for the wheel to slow down as 10.0 seconds.

  - Using the formula α = (ωf - ω0) / t, where ωf is the final angular speed and t is the time, we get α = (0 - 9.0) / 10.0 = -0.9 rotations per second squared.

2. Calculate the total number of rotations during deceleration:

  - Given that the wheel undergoes 65.0 complete rotations during this time, we divide the rotations by the initial angular speed to find the time it takes to complete these rotations:

  - Time = Rotations / Initial angular speed = 65.0 / 9.0 = 7.22 seconds.

3. Use the equation to find the time to stop (tstop):

  - We can use the formula ωf = ω0 + α * tstop, where ωf = 0 (as the wheel comes to a stop), ω0 = 9.0 rotations per second, and α = -0.9 rotations per second squared.

  - Substituting these values, we get 0 = 9.0 + (-0.9) * tstop.

  - Solving for tstop gives us tstop = 9.0 / 0.9 = 10 seconds.

So, it will take approximately 23.6 seconds for the bike wheel to come to a complete stop.

The problem involves angular motion and the effects of friction on a spinning bike wheel. To solve for the time it takes for the wheel to stop, we first determine the angular deceleration using the given initial angular speed and the time it takes for the wheel to slow down. With this deceleration, we calculate the time it takes for the wheel to complete the given number of rotations. Finally, using the formula for angular motion with constant acceleration, we find the additional time needed for the wheel to stop completely.

Complete Question:
A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10.0 s period following the inital spin, the bike wheel undergoes 65.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ t s will it take the bike wheel to come to a complete stop?

The wheel will take approximately 7.22 seconds to come to a complete stop.

To determine how much more time Δt it will take for the bike wheel to come to a complete stop, let's follow these steps:

Given:

Assuming constant angular deceleration (α),

We can find α using the formula:

Solving for α:

so

Hence,

Now, calculate the time (Δt) it will take to stop from 13π rad/s:

So, the additional time required for the wheel to stop, Δt, is approximately 7.22 seconds.

Answer:

1.47*10^5m/s

Explanation:

Using law of conservation of momentum

Since the uranium was initially at rest then the total momentum (did not change) is conserved and its initial value is zero ; 0= M1*V1+M2V2

-M1V1 = M2V2 where V1 is the velocity of the alpha particle and m1 is its mass and M2 is the mass of the thorium and V2 is its velocity.

- 6.64*10^-27*8.56*10^6 = 3.88*V2

V2 = -5.69*10^-20/3.88*10^-25 = -1.47*10^5m/s the negative sign mean the vector velocity of the nucleus point in opposite direction to the particle.

The recoil speed of the thorium nucleus after uranium-238 decays is approximately 0.14717 x 10^6 m/s. This is calculated using the law of conservation of momentum, where the initial momentum of the uranium-238 nucleus at rest is equal to the combined momentum of the emitted α-particle and the resulting thorium nucleus.

The recoil speed of the thorium nucleus can be calculated using the law of conservation of momentum. Before the decay, the momentum of the uranium-238 nucleus (the parent nuclide) is zero, as it's at rest. After the decay, the momentum has to remain zero for the system, including the α-particle and the daughter nucleus thorium-234.

It's given that the mass of the α-particle is 6.64×10−27 kg and it's emitted at a speed of 8.57×10^6 m/s. So, the momentum of the α-particle is the product of its mass and velocity, which is then equated to the momentum of the Thorium-234, which is its mass times its velocity (recoil speed).

Setting the two momentums equal to each other and solving for the recoil speed of thorium (V) gives us: (6.64×10−27 kg * 8.57×10^6 m/s) / 3.88 × 10−25 kg = V. Calculating this gives a recoil speed of about 0.14717 * 10^6 m/s for the thorium nucleus.

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Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis  ( tan^-1 (13.5 / 16.4)

Answer:

 [tex]v = 11.29\ m/s[/tex]

Explanation:

given,

bead is released from the height = 24.5 m from bottom

radius of the loop = 9 m

acceleration due to gravity = 9.8 m/s

A be the top most point of the loop

Difference of elevation from the top

     H = 24.5 - 2 x r

     H = 24.5 - 2 x 9

     H = 24.5 - 18

     H = 6.5 m

now using conservation of energy

   KE = PE

 [tex]\dfrac{1}{2}mv^2 = m g H[/tex]

 [tex]v^2 = 2 g H[/tex]

 [tex]v = \sqrt{2 g H}[/tex]

 [tex]v = \sqrt{2 \times 9.8 \times 6.5}[/tex]

 [tex]v = \sqrt{127.4}[/tex]

 [tex]v = 11.29\ m/s[/tex]

speed at point A is equal to  [tex]v = 11.29\ m/s[/tex]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

The direction of rotation of a proton can be determined using the right-hand rule in physics when considering the direction of its magnetic moment and the uniform magnetic field it's immersed in. The resultant rotation direction works to facilitate energy minimization in the system.

In the presence of an external magnetic field, the principle of physics concerning the interaction between magnetic fields and magnetic moments state that the magnetic moment of an object, in this case a proton, will align itself along the direction of the external magnetic field. This means, if the magnetic moment vector mu of the proton is directed along the positive z-axis, and the magnetic field is also along the positive z-axis, the proton's rotation will be in the direction opposite to that of the magnetic field following the right-hand rule. The proton's rotation and the magnetic field directions facilitate energy minimization in the system. Hence, the direction of rotation of a proton, based on the direction of its magnetic moment and the direction of the uniform magnetic field it is immersed in, can be determined by the right-hand rule; if your thumb points in the direction of the proton's magnetic moment (as opposed to the magnetic field), your fingers will curl in the direction of its rotation.

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Answer:

Height of slide = 0.936 m

Explanation:

Gravitational potential = Mass x Acceleration due to gravity x Height

P.E = mgh

Mass, m = 49 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = ?

Potential energy, P.E = 450 J

Substituting

              P.E = mgh

              450 = 49 x 9.81 x h

                 h = 0.936 m

Height of slide = 0.936 m

Answer:

Solids

Explanation:

Some of the properties that limit the motion of particles in a solid:

1. The molecules of a solid are very closely packed together thereby limiting their motion compare to molecules of liquid which are free to move  and gas with loosely bound molecules.

2. The molecules of a solid are held together by strong inter-molecular force due to very small inter-molecular spaces.

3. The motion of molecules in a solid vibrates only in their mean or fixed positions.

4. The particles are arranged in definite pattern and shape. A solid neither takes on the shape of its container like liquid, nor does it fill the entire volume available like a gas.

5. A solid is rigid and does not flow like liquid and gas.

6. A solid has a definite volume and can not be easily compressed.

Forces of attraction most limit the motion of particles in solids. This is because particles in a solid are tightly packed together and held in place by strong intermolecular forces. As solids change to liquids or gases, these forces lessen, allowing particles more freedom.

Forces of attraction most limit the motion of particles in solids. A solid's particles are tightly packed together and held in place by strong intermolecular forces. For example, atoms in a solid are always in close contact with neighboring atoms, held in place by forces (Figure 14.2 (a)). In contrast, particles in a liquid, while also in close contact, are able to slide over one another (b), and particles in a gas move about freely (c).

As solids are heated and change state to a liquid and then a gas, these forces of attraction lessen, allowing particles more freedom of movement. In summary, the state of a substance (solid, liquid, or gas) is largely dependent upon the behavior of its particles, which, in turn, is determined by the attractive forces between them. The stronger the forces of attraction, the less motion the particles will have, as is the case with solids.

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The tangential velocity of the car is 124.5 m/s

Explanation:

The centripetal acceleration of an object in uniform circular motion is given by:

[tex]a=\frac{v^2}{r}[/tex]

where

a is the acceleration

v is the tangential velocity

r is the radius of the circle

For the car in this problem, we have:

[tex]a=100 m/s^2[/tex] is the centripetal acceleration

r = 155 m is the radius of the turn

Solving for v, we find the tangential velocity of the car:

[tex]v=\sqrt{ar}=\sqrt{(100)(155)}=124.5 m/s[/tex]

Learn more about centripetal acceleration:

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Answer:Decrease Wavelength

Explanation:

To remove the Electrons from a metal surface the minimum amount of energy is provided i.e. threshold Energy is Provided to remove the electrons.

If no photoelectrons are emitted then it can be Concluded that Energy is less than threshold Energy.

Energy is inversely Proportional to the wavelength of light thus to increase Energy we have to decrease the wavelength of monochromatic emission.

a. The energy density of the magnetic field ([tex]\(u_B\)[/tex]) at the surface of the wire is [tex]\(3.08875 \times 10^{-7} \, \text{J/m}^3\)[/tex]

b. The energy density of the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire is [tex]\(3.74883 \times 10^{-3} \, \text{J/m}^3\).[/tex]

To calculate the energy density of the magnetic field ([tex]\(u_B\)[/tex]) and the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire, we first need to find the magnetic field strength (B) and the electric field strength (E) at that point.

Given:

- Current (I) = 11 A

- Wire diameter (d) = 2.8 mm = 2.8 x 10^-3 m

- Resistance per unit length (R) = 3.7 Ω/km = 3.7 x 10^3 Ω/m

We can calculate the radius (r) of the wire:

[tex]\[ r = \frac{d}{2} = \frac{2.8 \times 10^{-3}}{2} = 1.4 \times 10^{-3} \, \text{m} \][/tex]

(a) Magnetic Field Energy Density ([tex]\(u_B\)[/tex]):

The magnetic field strength at the surface of the wire (B) is given by Ampère's Law:

[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]

[tex]\[ B = \frac{4\pi \times 10^{-7} \times 11}{2\pi \times 1.4 \times 10^{-3}} \][/tex]

[tex]\[ B = \frac{4 \times 11 \times 10^{-7}}{2 \times 1.4 \times 10^{-3}} \][/tex]

[tex]\[ B = \frac{4 \times 11}{2 \times 1.4} \times 10^{-4} \][/tex]

[tex]\[ B = \frac{44}{2.8} \times 10^{-4} \][/tex]

[tex]\[ B = 15.714 \times 10^{-4} \][/tex]

[tex]\[ B = 1.571 \times 10^{-3} \, \text{T} \][/tex]

The energy density of the magnetic field ([tex]\(u_B\)[/tex]) is given by:

[tex]\[ u_B = \frac{B^2}{2\mu_0} \][/tex]

[tex]\[ u_B = \frac{(1.571 \times 10^{-3})^2}{2 \times 4\pi \times 10^{-7}} \][/tex]

[tex]\[ u_B = \frac{2.471 \times 10^{-6}}{8\pi \times 10^{-7}} \][/tex]

[tex]\[ u_B = \frac{2.471}{8} \times 10^{-6} \][/tex]

[tex]\[ u_B = 0.308875 \times 10^{-6} \][/tex]

[tex]\[ u_B = 3.08875 \times 10^{-7} \, \text{J/m}^3 \][/tex]

(b) Electric Field Energy Density ([tex]\(u_E\)[/tex]):

The electric field strength (E) at the surface of the wire is given by Ohm's Law:

[tex]\[ E = \frac{V}{r} \][/tex]

[tex]\[ E = \frac{IR}{r} \][/tex]

[tex]\[ E = \frac{11 \times 3.7 \times 10^3}{1.4 \times 10^{-3}} \][/tex]

[tex]\[ E = \frac{40.7 \times 10^3}{1.4 \times 10^{-3}} \][/tex]

[tex]\[ E = \frac{40.7}{1.4} \times 10^3 \][/tex]

[tex]\[ E = 29.071 \times 10^3 \][/tex]

[tex]\[ E = 29.071 \times 10^3 \, \text{V/m} \][/tex]

The energy density of the electric field ([tex]\(u_E\)[/tex]) is given by:

[tex]\[ u_E = \frac{\varepsilon_0 E^2}{2} \][/tex]

[tex]\[ u_E = \frac{8.85 \times 10^{-12} \times (29.071 \times 10^3)^2}{2} \][/tex]

[tex]\[ u_E = \frac{8.85 \times 10^{-12} \times 845.96 \times 10^6}{2} \][/tex]

[tex]\[ u_E = \frac{7.49766 \times 10^{-3}}{2} \][/tex]

[tex]\[ u_E = 3.74883 \times 10^{-3} \][/tex]

So, the energy density of the magnetic field ([tex]\(u_B\)[/tex]) at the surface of the wire is [tex]\(3.08875 \times 10^{-7} \, \text{J/m}^3\)[/tex] and the energy density of the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire is [tex]\(3.74883 \times 10^{-3} \, \text{J/m}^3\).[/tex]

Answer:

m2 = 0·52 kg

Explanation:

As the pulley is massless and frictionless the tension in the string on both sides of the pulley will be the same

Given that both have same acceleration a = 0·5 m/s²

Forces acting on mass m1 is tension and force of gravity

You can observe the direction of forces acting on two blocks in the file attached

And in the file attached the force of gravity acting on mass m2 is resolved into two perpendicular components of which one acts along the wedge and the other perpendicular to the wedge

Let the tension in the string be T N

And the frictional force acting on mass m2 is μ×N as sliding is taking place

By applying Newton's second law to the block of mass m1

m1×g - T = m1 × a

⇒ T = m1×(g - a)

⇒T = 0·5×(9·8-0·5)

⇒ T= 4·65 N

Let the normal reaction acting on mass m2 be N

By applying Newton's second law to the block of mass m2 in the direction perpendicular to the wedge

we get

N = m2×g×cosθ

By applying Newton's second law to the block of mass m2 along the wedge

T -  μ×N - m2×g×sinθ = m2×a

Substitute N =  m2×g×cosθ in the above equation

T - μ×m2×g×cosθ - m2×g×sinθ = m2×a

⇒ T = m2 × ( μ×g×cosθ + g×sinθ + a)

By the substituting the corresponding values

4·65 = m2 × 8·8

⇒ m2 = 0·52 kg

 The mass of block 2, m2.is mathematically given as

m2 = 0·52 kg

Question Parameter(s):

Generally, the equation for the  Newton's second law is mathematically given as

m1×g - T = m1 × a

Therefore

T = m1*(g - a)

T = 0·5*(9·8-0·5)

T= 4·65 N

Therefore for

T -  u*N - m2*g×sin[tex]\theta[/tex] = m2*a

4·65 = m2 × 8·8

m2 = 0·52 kg

In conclusion,  the mass of block 2

m2 = 0·52 kg

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Answer:

1. True

2. False

3. True

Explanation:

Semiconductors are the materials whose conduction is intermediary to that of the conductor and insulator and can be made more conducting by mixing impurities or we call it as doping.

In semiconductors, the concentration of the holes in the valence band and the electrons in the conduction band is usually very small and can be varied noticeably by doping.

(1) After being excited by a photon, an electron can move to the conduction band as a hole is created in the valence band band and this hole moves to the p-side of the p-n junction.

(2) It is not the n-type but the p-type semiconductor with missing electrons or holes in the valence band of lower energy.

(3) Electrons are free to move in a partially filled conduction band and valence band.

There is no movement of electrons along the band which are completely full or totally vacant.

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = [tex]11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}[/tex]

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = [tex]390\ W/m.^{\circ}C[/tex]

Conduction of heat from the rod per second is given by:

[tex]q = \frac{KA\Delta T}{L}[/tex]

where

[tex]\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C[/tex] = temperature difference between the two ends of the rod.

Thus

[tex]q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s[/tex]

Now,

To calculate the mass, M of the ice melted per sec:

[tex]M = \frac{q}{L_{w}}[/tex]

where

[tex]L_{w}[/tex] = Latent heat of fusion of water = 333 kJ/kg

[tex]M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g[/tex]

To find the number of grams of ice melted per second, we can calculate the rate of heat transfer through the copper rod from the boiling water to the ice-water mixture using Fourier's Law of Heat Conduction. The formula is:

[tex]\[ Q = kA \frac{\Delta T}{L} \][/tex]

Where:

- Q is the rate of heat transfer (in watts, W)

- k  is the thermal conductivity of the material (in W/m°C)

-  A  is the cross-sectional area of the rod (in m²)

- [tex]\( \Delta T \)[/tex] is the temperature difference between the hot and cold ends (in °C)

- L is the length of the rod (in meters)

First, we need to find the temperature difference [tex]\( \Delta T \)[/tex]between the hot end (boiling water) and the cold end (ice-water mixture). We know water boils at 100°C and ice-water mixture is at 0°C. Therefore,[tex]\( \Delta T = 100°C - 0°C = 100°C \).[/tex]

Given:

- Cross-sectional area [tex]\( A = 11.6 \, \text{cm}^2 = 0.00116 \, \text{m}^2 \)[/tex]

- Length of the rod [tex]\( L = 19.6 \, \text{cm} = 0.196 \, \text{m} \)[/tex]

- Thermal conductivity of copper [tex]\( k = 390 \, \text{W/m°C} \)[/tex]

- Latent heat of fusion of ice [tex]\( L_f = 334 \, \text{J/g} \)[/tex]

Now, let's calculate the rate of heat transfer  Q :

[tex]\[ Q = (390 \times 0.00116) \times \frac{100}{0.196} \][/tex]

[tex]\[ Q \approx 236.8 \, \text{W} \][/tex]

Next, we need to convert this rate of heat transfer into grams of ice melted per second using the latent heat of fusion of ice:

[tex]\[ \text{Grams of ice melted per second} = \frac{Q}{L_f} \][/tex]

[tex]\[ \text{Grams of ice melted per second} = \frac{236.8 \, \text{J/s}}{334 \, \text{J/g}} \][/tex]

[tex]\[ \text{Grams of ice melted per second} \approx 0.708 \, \text{g/s} \][/tex]

Therefore, approximately [tex]\( 0.708 \)[/tex] grams of ice melt each second. This means that [tex]\( 0.708 \)[/tex] grams of ice in the ice-water mixture will melt every second due to the heat transferred from the boiling water through the copper rod.

Answer:

option D.

Explanation:

given,                                                                          

Average rainfall in Minnesota = 50 cm/ year

area of the landfill = 5000 m²            

60% of the water runs off the landfill.

efficiency of leachate collection = 80%

volume of leachate treated per year = ?

volume of rainfall in a year                  

        = 0.5 x 5000                  

        = 2500 m³                    

Volume of  water infiltrated    

     = 40 % of volume of rainfall

     = 0.4 x  2500                  

    = 1000 m³                    

Volume of leachate per year

    = 80 % of 1000              

    = 0.8 x 1000                  

    = 800 m³                  

The correct answer is option D.

Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by  [tex]\frac{\pi}{2}[/tex]. Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by [tex]\frac{\pi}{2}[/tex].

Answer:

(A) 0.2306 m

(B) 1.467 Hz

(C) 0.1152 m

Explanation:

spring constant (K) = 16.4 N/m

mass (m) = 0.193 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) force = Kx,  where x = extension

   mg = Kx

   0.193 x 9.8 = 16.4x

   x = 0.1153 m

  now the mass actually falls two times this value before it gets to its equilibrium position ( turning  point ) and oscillates about this point

therefore

2x = 0.2306 m

(B) frequency (f) = \frac{1}{2π} x [tex]\sqrt{\frac{k}{m}}[/tex]

     frequency (f) = \frac{1}{2π} x [tex]\sqrt{\frac{16.4}{0.193}}[/tex]

     frequency = 1.467 Hz  

(C) the amplitude is the maximum position of the mass from the equilibrium position, which is half the distance the mass falls below the initial length of the spring

= \frac{0.2306}{2} =  0.1152 m

Final answer:

The mass descends by 0.1154 meters, the frequency of SHM is about 1.83 Hz, and the amplitude of the oscillation is 0.1154 meters.

Explanation:

When a mass of 0.193 kg is hung on a vertical massless spring of spring constant 16.4 N/m, the spring is stretched due to the weight of the mass. The distance that the body descends is equal to the distance at which the spring force balances the gravitational force on the mass. This can be calculated by using Hooke's Law (F = kx), where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. The force due to gravity on the mass is given by W = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s2).

(a) To find the descent (x), set the spring force equal to the gravitational force: mg = kx, which yields x = mg/k. Plugging in the values yields x = (0.193 kg)(9.8 m/s2) / 16.4 N/m, which gives x = 0.1154 meters.

(b) The frequency of the resulting SHM (Simple Harmonic Motion) can be determined by the formula f = (1/2π) √(k/m). Substituting the values, f = (1/2π) √(16.4 N/m / 0.193 kg), which gives f ≈ 1.83 Hz.

(c) The amplitude of SHM is equal to the maximum displacement from the equilibrium position, which is the same as the distance calculated in part (a), hence the amplitude is 0.1154 meters.

Answer:

The satellite has a period of 204.90 days

Explanation:

The period can be determine by means of Kepler's third law:

[tex]T^{2} = r^{3}[/tex]  (1)

Where T is the period of revolution and r is the radius.

[tex]\sqrt{T^{2}} = \sqrt{r^{3}}[/tex]

[tex]T = \sqrt{r^{3}}[/tex] (2)

The distance between the moon and the Earth has a value of 384400 km, therefore:

[tex]r = (384400km)*(0.50)[/tex]

[tex]r = 192200 km[/tex]

Finally, equation 2 can be used:

[tex]T = \sqrt{(192200)^{3}}[/tex]

[tex]T = 84261672 km[/tex]

However, the period can be expressed in days, to do that it is necessary to make the conversion from kilometers to astronomical units:

An astronomical unit (AU) is the distance between the Earth and the Sun ([tex]1.50x10^{8} km[/tex])

[tex]T = 84261672 km \cdot \frac{1AU}{1.50x10^{8} km}[/tex] ⇒ [tex]0.561 AU[/tex]

But 1 year is equivalent to 1 AU according to Kepler's third law, since 1 year is the orbital period of the Earth.

[tex]T = 0.561 AU \cdot \frac{1year}{1AU}[/tex] ⇒ [tex]0.561 year[/tex]

[tex]T = 0.561 year \cdot \frac{365.25 days}{1year}[/tex] ⇒ [tex]204.90 days[/tex]

[tex]T = 204.90 days[/tex]

Hence, the satellite has a period of 204.90 days.

Final answer:

The period of the satellite's orbit around the Earth is approximately 5059 seconds.

Explanation:

The period of a satellite orbiting around the Earth depends on the radius of its orbit. According to Kepler's third law, the period is related to the radius by the equation T^2 = (4π²r^3) / (GM), where T is the period, r is the radius, G is the gravitational constant, and M is the mass of the Earth.

In this case, the radius of the satellite's orbit is one half the distance from the Earth to the Moon, which is half of 3.84 × 10^8 meters. So, the radius is 1.92 × 10^8 meters. Plugging this value into the equation, we can calculate the period of the satellite's orbit.

Solving for T, we get:

T = 2π√[(r^3) / (GM)]

Substituting the values, we have:

T = 2π√[(1.92 × 10^8)^3 / (6.67 × 10^-11 × 5.97 × 10^24)]

T = 2π√(6.71 × 10^21 / 3.95 × 10^35)

T ≈ 5059 seconds

So, the period of the satellite's orbit is approximately 5059 seconds.

Fnet = ma = m*0.23g = mgsinΘ - Ff

where Ff is the friction force upslope.

net torque τ = Ff * r,

but also τ = I*α =(2/3)mr² * a/r = 2mra/3 = 2mr(0.23g)/3 = 0.46mrg/3

Then Ff * r = 0.46mrg / 3

Ff = 0.46mg/3 → put this into net force equation:

m*0.23g = mgsinΘ - 0.46mg/3 → mg cancels; multiply through by 3

0.69 = 3sinΘ - 0.46

3sinΘ = 1.15

sinΘ = 1.15/3

Θ = arctan(1.15/3) = 22.54 º

Answer:

Explanation:

The photons travel faster faster through space because photons always travel through space faster than electrons, in fact when an electron gets hitted by a photon this boost its speed

Answer:

I would think the answer is color, if the wavelength is within the visible light spectrum. This could be answered in different ways but I'm pretty sure the answer you are looking for is hue/color.

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J      

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy ([tex]E_{k}[/tex]) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

                                        [tex]E_{k}[/tex] = h(f − f₀)

                                        [tex]E_{k}[/tex] = hf - hf₀

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

                                        [tex]E_{k}[/tex] = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy [tex]E_{k}[/tex] = 4.16×10⁻¹⁷ J  

Speed of light  c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js                                

                                        E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

                                        E = 53.8356 x 10⁻¹⁶ J

from [tex]E_{k}[/tex] = E - ϕ ;

                                        ϕ = E - [tex]E_{k}[/tex]

                                        ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

                                        ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J      

Answer:

A=60.06 x 10⁻⁵ m²

Explanation:

Given that

Weight of the person = 864 N

Weight of the bike = 85 N

The total weight ,car + bike ,Wt= 864 + 85 N

 Wt = 949 N

2 p = 949 N

The weight on the each tire(F) = 474.5  N

P = 7.9 x 10 ⁵  Pa

We know that

Force = Pressure x Area

F= P A

By putting the values

474.5 =  7.9 x 10 ⁵  A

A=60.06 x 10⁻⁵ m²

Therefore area of contact A

A=60.06 x 10⁻⁵ m²

Answer:

electrostatic attraction.

Explanation:

As we know those atom donate the electron gets positively charged and those atoms gains the electron gets negatively charged.The charge on the atoms are the responsible for the force in the ionic bond.

We know that force between two charge particle is known as electrostatics force.If charge particle is having same sign charge then it will be repulsive force and If charge particle is having opposite sign charge then it will be attraction  force.

Therefore force in the ionic solid is electrostatic attraction.

Final answer:

Ionic solids are primarily held together by electrostatic attraction, with dispersion forces playing a possible but minor role. Metallic bonds, covalent bonds, dipole-dipole forces, and hydrogen bonds do not typically contribute to the structure of ionic solids.

Explanation:

The kinds of forces that hold ionic solids together include electrostatic attraction and possibly dispersion forces if there are also nonpolar parts or induced dipoles present. Electrostatic attraction, also known as ionic bonding, is the main force that holds together the ions in an ionic solid, resulting from the attraction between positively charged cations and negatively charged anions. However, dispersion forces can also play a lesser role when nonpolar regions or molecules are adjacent to the ionic components, though they are not the primary force in ionic solids.

Metallic bonds, covalent bonds, dipole-dipole forces, and hydrogen bonds are not typically involved in the structure of ionic solids. Metallic bonds are specific to metal lattices, covalent bonds connect atoms within molecules or network solids, dipole-dipole forces and hydrogen bonds are types of intermolecular forces mainly significant in molecular compounds, not ionic compounds.

Answer:

Circuit breaker

Explanation:

Circuit breaker is the devise designed to protect the circuit from over current by opening the circuit automatically. Breaker can also be off manually by toggle switch. Earlier fuses were used but circuit breakers have replaced them. Fuse and circuit breakers operates differently. in case of overloading fuses blown off and opens the circuit while circuit breaker opens the circuit automatically without being blown off.

Final answer:

A circuit breaker is a safety device in a circuit that opens the circuit when an overcurrent occurs. It is faster than fuses and can be reset.

Explanation:

A circuit breaker is a device designed to open and close a circuit by non-automatic means and to open the circuit automatically on a predetermined overcurrent without damage to itself when properly applied within its rating. It acts as a safety device that switches off an appliance if the current in the circuit is too strong.

Circuit breakers are rated for a maximum current and can be reset, reacting much faster than fuses. They consist of components like bimetallic strips that respond to heat to break the electrical connection in the circuit.

The partial pressure of ozone at 441 ppb with an atmospheric pressure of 0.67 atm is [tex]2.947 x 10^-7 atm.[/tex]

Atmospheric pressure=0.67 atm

We have to calculate the partial pressure of ozone at 441 ppb .

To calculate the partial pressure of ozone at 441 ppb, we need to use the formula:

Partial Pressure = Concentration x Total Pressure

Given that the atmospheric pressure is 0.67 atm and the concentration of ozone is 441 ppb (441 parts per billion), we can calculate the partial pressure as follows:

Partial Pressure of Ozone

= (441/1,000,000,000) x 0.67 atm

[tex]= 2.947 x 10-7[/tex]

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