A current-carrying wire passes through a region of space that has a uniform magnetic field of 0.92 T. If the wire has a length of 2.6 m and a mass of 0.60 kg, determine the minimum current needed to levitate the wire. A

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

   Volume = 470 cm³

   Density = 3.55 g/cm³

   Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

   Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

   Volume = 470 cm³

   Density of liquid = 1 g/cm³

   [tex]\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N[/tex]

c) Mass of the water displaced = Volume of body x Density of liquid

   Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

   Weight of rock under water = 16.37 - 4.61  =11.76 N

Answer:

542.06 K

Explanation:

v = rms speed of oxygen molecule = 650 m/s

M = molecular mass of the oxygen molecule = 32 g = 0.032 kg

R = universal gas constant = 8.314 J/(mol K)

T = temperature of the gas

Rms speed of oxygen molecule is given as

[tex]v = \sqrt{\frac{3RT}{M}}[/tex]

[tex]650 = \sqrt{\frac{3(8.314)T}{0.032}}[/tex]

T = 542.06 K

Answer:

[tex]f = +42.9 cm[/tex]

[tex]P =+2.33Dioptre[/tex]

Explanation:

As we know that Far sighted person has near point shifted to 60 cm distance

so he is able to see the object 60 cm

and the person want to see the objects at distance 25 cm

so here the image distance from lens is 60 cm and the object distance from lens is 25 cm

now from lens formula we have

[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]

[tex]-\frac{1}{60} + \frac{1}{25} = \frac{1}{f}[/tex]

[tex]f = +42.9 cm[/tex]

Now we know that power of lens is given as

[tex]P = \frac{1}{f}[/tex]

[tex]P = \frac{1}{0.429} = +2.33Dioptre[/tex]

Explanation:

It is given that,

Length of solenoid, l = 28 cm = 0.28 m

Area of cross section, A = 0.475 cm² = 4.75 × 10⁻⁵ m²

Current, I = 85 A

(a) The inductance of the solenoid is given by :

[tex]L=\dfrac{\mu_oN^2A}{l}[/tex]

[tex]L=\dfrac{4\pi\times 10^{-7}\times (645)^2\times 4.75\times 10^{-5}}{0.28}[/tex]

L = 0.0000886 H

[tex]L=8.86\times 10^{-5}\ H[/tex]

(b) Energy contained in the coil's magnetic field is given by :

[tex]U=\dfrac{1}{2}LI^2[/tex]

[tex]U=\dfrac{1}{2}\times 8.86\times 10^{-5}\ H\times (85\ A)^2[/tex]

U = 0.3200675 Joules

or

U = 0.321 Joules

Hence, this is the required solution.

Answer:

The answer is 1.1A

Explanation:

See the attached file

Answer:

1.34 x 10^3 Pa

Explanation:

density of oil = 0.85 x 10^3 kg/m^3

g = 9.81 m/s^2

height of oil column = 16.1 cm = 0.161 m

Pressure on the surface of water = height of oil column x density of oil x g

                                                      = 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa

Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

Answer:

168.32 mile

Explanation:

1 mile = 1.61 km

1.61 km = 1 mile

1 km = 1 / 1.61 mile

So, 271 km = 271 / 1.61 = 168.32 mile

Answer:

[tex]KE_f = 372 J[/tex]

Explanation:

The forces on the box while it is sliding down are

1). Component of the weight along the inclined plane

2). Friction force opposite to the motion of the box

3). Applied force on the box

now we know that component of the weight along the inclined plane is given as

[tex]F_g = mgsin\theta[/tex]

[tex]F_g = (20.0)(9.8)sin30 = 98 N[/tex]

Now we know that other component of the weight of object is counterbalanced by the normal force due to inclined plane

[tex]F_n = mgcos\theta[/tex]

[tex]F_n = (20.0)(9.8)cos30 = 170 N[/tex]

now the kinetic friction force on the box is given as

[tex]F_k = \mu F_n[/tex]

[tex]F_k = 0.100(170) = 17 N[/tex]

now the Net force on the box which is sliding down is given as

[tex]F_{net} = F_g - F_k - F_{applied}[/tex]

[tex]F_{net} = 98 - 17 - 50 = 31 N[/tex]

now the work done by net force = change in kinetic energy of the box

[tex]F.d = KE_f - KE_i[/tex]

[tex]31(12) = KE_f - 0[/tex]

[tex]KE_f = 372 J[/tex]

Final answer:

The increase in the kinetic energy of the box is calculated by finding the net work done on the box as it slides down the incline, which is found to be 373.2 J. The closest answer from the options provided is 372 J.

Explanation:

We need to calculate the increase in the kinetic energy of the box as it slides down the incline. First, let's determine the forces acting on the box:

Gravitational force component along the incline: Fg = m*g*sin(θ) = 20.0 kg * 9.81 m/s2 * sin(30°) = 98.1 N

Kinetic friction force: [tex]F_{k}[/tex] = μk * N = μk * m*g*cos(θ) = 0.100 * 20.0 kg * 9.81 m/s2 * cos(30°) = 17.0 N

Applied force up the incline: Fa = 50.0 N

Next, calculate the net force on the box:

Fnet = Fg - [tex]F_{k}[/tex] - Fa = 98.1 N - 17.0 N - 50.0 N = 31.1 N

Now, calculate the work done by the net force, which equals the increase in kinetic energy:

Work = Fnet * d = 31.1 N * 12.0 m = 373.2 J

The closest answer to this calculated value is 372 J.

Answer:

Option D is the correct answer.

Explanation:

Young's modulus is the ratio of tensile stress and tensile strain.

Bulk modulus is the ratio of pressure and volume strain.

Rigidity modulus is the ratio of shear stress and shear strain.

Here we are asked about relationship between elastic shear stress and shear strain. We have rigidity modulus is the ratio of shear stress and shear strain.

Option D is the correct answer.

The relationship between elastic shear stress and shear strain is characterized by the modulus of rigidity or shear modulus, which is a key property in understanding a material's deformation under shear forces.

The relationship between elastic shear stress and shear strain is represented by the modulus of rigidity, also known as the shear modulus. This modulus is a type of elastic modulus specific to shear stress and is the proportionality constant that relates shear stress to shear strain within the linear elastic region of a material's response to stress, as described by Hooke's Law.

There are different types of elastic modulus for different types of stress and strain; for tensile stress, the modulus is known as Young's modulus, for bulk stress, it is the bulk modulus, and for shear stress - which is the focus of this question - it is the shear or rigidity modulus. The modulus of rigidity is crucial for determining how a material will deform under shear forces and is a fundamental property used in engineering and construction to ensure materials behave as expected under load.

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]

now we have

[tex]I = mk^2[/tex]

here k = radius of gyration of object

also for pure rolling we have

[tex]v = R\omega[/tex]

so now we will have

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})[/tex]

[tex]mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})[/tex]

[tex]v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}[/tex]

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

[tex]mk^2 = mR^2[/tex]

k = R

For spherical shell

[tex]mk^2 = \frac{2}{3}mR^2[/tex]

[tex]k = \sqrt{\frac{2}{3}} R[/tex]

For solid sphere

[tex]mk^2 = \frac{2}{5}mR^2[/tex]

[tex]k = \sqrt{\frac{2}{5}} R[/tex]

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

The equivalent resistance of the combination is 15 ohm. Option C is correct.

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

Case 1: Two 20 ohm resistors are connected in parallel;

[tex]\rm \frac{1}{R_{eq}_1} =\frac{1}{R_1} +\frac{1}{R_2} \\\\ \rm \frac{1}{R_{eq}_1} =\frac{1}{20} +\frac{1}{20} \\\\ R_{eq}_1}= 10 \ ohm[/tex]

Case 2: Two 10-ohm resistors are connected in parallel.

[tex]\rm \frac{1}{R_{eq}_2} =\frac{1}{R_1} +\frac{1}{R_2} \\\\ \rm \frac{1}{R_{eq}_2} =\frac{1}{10} +\frac{1}{10} \\\\ R_{eq}_2}= 5 \ ohm[/tex]

Case 3; Two combinations are connected in series.

[tex]\rm R= R_{eq_1}+R_{eq_2}\\\\ \rm R= 10+ 5 \\\\ R=15 \ ohm[/tex]

The equivalent resistance of the combination is 15 ohm.

Hence, option C is correct.

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Final answer:

Two 20 ohm resistors in parallel have an equivalent resistance of 10 ohms, two 10 ohm resistors in parallel have an equivalent resistance of 5 ohms. When these two combinations are connected in series, the total equivalent resistance is the sum of both, which is c) 15 ohms.

Explanation:

To find the equivalent resistance of the given combinations of resistors, we first need to understand how resistors combine in parallel and in series.

When resistors are in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances:

Req = 1 / (1/R1 + 1/R2)

For two 20 ohm resistors in parallel:

Req = 1 / (1/20 + 1/20) = 1 / (2/20) = 10 ohms

For two 10 ohm resistors in parallel:

Req = 1 / (1/10 + 1/10) = 1 / (2/10) = 5 ohms

When we have resistors in series, their resistances simply add up:

Rtotal = Req1 + Req2

So, for the 10 ohm equivalent and the 5 ohm equivalent resistors in series:

Rtotal = 10 ohms + 5 ohms = 15 ohms

Therefore, the correct answer to the student's question is (c) 15 Ohm.

Answer:

[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

[tex]\vec F_e = q\vec E[/tex]

[tex]\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3[/tex]

[tex]\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j[/tex]

Now force on moving charge due to magnetic field is given as

[tex]\vec F_b = q(\vec v \times \vec B)[/tex]

[tex]\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )[/tex]

[tex]\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i[/tex]

[tex]\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k[/tex]

Now net force due to both

[tex]F = F_e + F_b[/tex]

[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]

An electric and magnetic field exerts force on a proton moving with velocity in that field. The total force can be calculated from the Lorentz Force equation, which requires knowledge of the charge of the proton, its velocity, and the electric and magnetic fields it is experiencing.

In Physics, the total force acting on a charged particle moving through an electric field E and a magnetic field B is given by the Lorentz Force equation: F = q(E + v × B), where q is the charge of the particle, and v is its velocity.

By given that the proton's charge is q = 1.6×10^-19 C and proton's velocity v = (6.6i+2.8j-4.8k)x10^3 m/s, electric field E = (3.3i-4.5j)x10^3 V/m, and magnetic field B = (0.64i+0.40j)T, we can plug these values into the equation.

To find the cross product of v and B, we use the determinant of a 3x3 matrix. The value for the F vector can be calculated as follows: F = q [E + v × B] and the cross product 'v × B' is calculated as the determinant of a 3x3 matrix. This yields force values that are expected to be in i, j, and k components. These calculations need to be done carefully to ensure accuracy, but are straightforward with the use of any standard physics formula sheet.

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Answer:

346m/s

Explanation:

v1=335m/s

f1=200hz

f2=207hz

v2=?

v1/f1=v2/f2

335/200=v2/207

v2=335*207/200

v2=346m/s

Answer:

The conductivity of the wire is [tex]3.18\times10^{7}\ ohm-m[/tex].

Explanation:

Given that,

Diameter = 0.2 cm

Current = 20 A

Power = 4 W/m

We need to calculate the conductivity

We know that,

[tex]\sigma = \dfrac{1}{\rho}[/tex]

Using  formula of resistance

[tex]R = \dfrac{\rho l}{A}[/tex]....(I)

Where,

[tex]\rho[/tex] = resistivity

A = area

l = length

Using formula of power

[tex]P = i^2 R[/tex]

[tex]R = \dfrac{P}{i^2}[/tex]

Put the value of R in equation (I)

[tex]\dfrac{P}{i^2}=\dfrac{\rho l}{\pi r^2}[/tex]

[tex]\rho=\dfrac{P\pi r^2}{l\timesi^2}[/tex]

[tex]\sigma=\dfrac{l\times i^2}{P\pi r^2}[/tex]

Put the all values into the formula

[tex]\sigma=\dfrac{1\times(20)^2}{4\times3.14\times(0.1\times10^{-2})^2}[/tex]

[tex]\sigma=3.18\times10^{7}\ ohm-m[/tex]

Hence, The conductivity of the wire is [tex]3.18\times10^{7}\ ohm-m[/tex].

To solve for the minimum allowable conductivity of the wire, we calculate the resistance based on the maximum power dissipation and then rearrange the formula for resistance to solve for conductivity. The minimum allowable conductivity for the wire is 3.18
* 10^4 (
Ω
m)^-1.

Minimum Allowable Conductivity of the Wire

To find the minimum allowable conductivity of a 0.2 cm diameter wire carrying a 20-A current with a maximum power dissipation of 4W/m, we first need to calculate the resistance of a 1-meter length of the wire based on the power dissipation.

Since power dissipation (
P) along the wire is given by P = I^2
* R, where I is the current and R is the instantaneous resistance of the wire, we can rearrange to find R = P / I^2. Substituting the given values, we get R = 4W / (20A)^2 = 0.01
Ω/m.

The resistance R of a conductor is also given by R =
L / (
σ
* A), where L is the length, σ (sigma) is the conductivity, and A is the cross-sectional area. We can rearrange and solve for the conductivity σ = L / (R
* A).

For a wire with a diameter of 0.2 cm, the cross-sectional area A is
π
* (0.1 cm)^2. Converting to meters, A = 3.14 * (0.001 m)^2 = 3.14
* 10^-6 m^2.

Now, substitute L = 1 m, R = 0.01 Ω/m, and A = 3.14
* 10^-6 m^2 into the formula to find the conductivity σ. The minimum allowable conductivity can then be calculated as σ = 1m / (0.01 Ω/m
* 3.14
* 10^-6 m^2) which yields σ = 3.18
* 10^4 (
Ω
*m)^-1.

Answer:

66.5 m/s

Explanation:

m = mass of the golf ball = 0.031 kg

F = magnitude of force applied to the ball = 6240 N

Acceleration experienced by the ball is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{6240}{0.031}[/tex]

a = 201290.32 m/s²

d = distance for which the ball is in contact with the golf club = 0.011 m

v₀ = initial speed of the ball = 0 m/s

v = final speed of the ball = 0 m/s

Using the kinematics equation

v² = v₀² + 2 a d

v² = 0² + 2 (201290.32) (0.011)

v = 66.5 m/s

Answer:

[tex]F_2 = 1.10 \mu N[/tex]

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now since it depends inverse on the square of the distance so we can say

[tex]\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}[/tex]

now we know that

[tex]r_2 = 18.2 mm[/tex]

[tex]r_1 = 12.2 mm[/tex]

also we know that

[tex]F_1 = 2.45 \mu N[/tex]

now from above equation we have

[tex]F_2 = \frac{r_1^2}{r_2^2} F_1[/tex]

[tex]F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)[/tex]

[tex]F_2 = 1.10 \mu N[/tex]

Answer:

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

                             =  1000*0.350 = 350 kg

F  = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2  

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom  = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

The resistance of a wire is given by :

[tex]R=\rho\dfrac{l}{A}[/tex]

Where

[tex]\rho[/tex] is the resistivity of the wire

l = initial length of the wire

A = initial area of cross section

If length and the area of cross section of the wire is doubled then new length is l' and A', l' = 2 l and A' = 2 A

So, new resistance of the wire is given by :

[tex]R'=\rho\dfrac{l'}{A'}[/tex]

[tex]R'=\rho\dfrac{l}{A}[/tex]

R' = R

So, the resistance of the wire remains the same on doubling the length and the area of wire.

Answer:

8.475 x 10^7 Pa

Explanation:

h = 8648 m, g = 9.8 m/s^2, density of water, d = 1000 kg/m^3

Pressure at a depth is defined as the product of depth of water , acceleration due to gravity and density of water.

P = h x d x g

P = 8648 x 1000 x 9.8 = 8.475 x 10^7 Pa

Answer:

14.82 Volts

Explanation:

[tex]V_{back}[/tex] = Back emf of the motor

[tex]V_{battery}[/tex] = battery voltage = 17.2 Volts

[tex]i_{locked}[/tex] = Current in locked condition = 12.3 A

R = resistance

In locked condition, using ohm's law

[tex]R = \frac{V_{battery}}{i_{locked}}[/tex]

[tex]R = \frac{17.2}{12.3}[/tex]

R = 1.4 Ω

[tex]i_{normal}[/tex] = Current in normal condition = 1.7 A

Back emf of the motor is given as

[tex]V_{back}[/tex] = [tex]V_{battery}[/tex] - [tex]i_{normal} R[/tex]

[tex]V_{back}[/tex] = 17.2 - (1.7 x 1.4)

[tex]V_{back}[/tex] = 14.82 Volts

Final answer:

The angular momentum of a particle about the origin is given by the cross product of its position vector and linear momentum. In this case, the position vector of the particle is (6.00 î + 5.70 t ĵ) and the mass of the particle is 1.70 kg. To find the angular momentum, calculate the linear momentum and take the cross product with the position vector.

Explanation:

The angular momentum of a particle about the origin is given by the cross product of its position vector and linear momentum. In this case, the position vector of the particle is represented by r = (6.00 î + 5.70 t ĵ) and the mass of the particle is 1.70 kg. To find the angular momentum, we need to calculate the linear momentum first.

The linear momentum of a particle is given by the product of its mass and velocity. The velocity vector is given by the derivative of the position vector with respect to time, which in this case is v = (0 î + 5.70 ĵ) m/s. Substituting the values, we can find the linear momentum, which is p = (1.70 kg)(0 î + 5.70 ĵ) m/s.

To find the angular momentum, we take the cross product of the position vector and linear momentum:

ŕ × p = (6.00 î + 5.70 t ĵ) × (1.70 kg)(0 î + 5.70 ĵ) m/s = (0 î + 34.65 î t + 9.69 ĵ) kg·m²/s

Therefore, the angular momentum of the particle about the origin as a function of time is (0 î + 34.65 î t + 9.69 ĵ) kg·m²/s.

The angular momentum of the particle about the origin as a function of time is L(t) = 58.14 t k kg·m²/s. This result was obtained by calculating the cross product of the position and linear momentum vectors.

To determine the angular momentum of a particle about the origin as a function of time, we start by using the given position vector r(t) = (6.00 î + 5.70t ĵ) in meters.

The linear momentum p of the particle is given by p = m v, where m is the mass and v is the velocity. Since mass m = 1.70 kg, we first need the velocity. The velocity v(t) is the derivative of the position vector:

v(t) = d(r(t))/dt = 0 î + 5.70 ĵ = 5.70 ĵ m/s

Now, the linear momentum p(t) is:

p(t) = m v(t) = 1.70 kg * 5.70 ĵ m/s = 9.69 ĵ kg·m/s

The angular momentum L about the origin is given by L = r × p. Performing the cross product calculation:

r = 6.00 î + 5.70t ĵ

p = 9.69 ĵ

r × p = (6.00 î + 5.70t ĵ) × (9.69 ĵ)

Calculating the cross product, we get:

Thus, the angular momentum as a function of time is:

L(t) = 58.14 t k kg·m²/s

Answer:

Major term is 'things that provide intense gravity'

Minor term is 'extremely dense objects'

Middle term is 'neutron stars'

Explanation:

Answer:

a) [tex]4.362580048\times 10^{-19}\ Joule[/tex]

b)[tex]0.27566898\times 10^{-19}\ Joule[/tex]

Explanation:

a) When, wavelength=λ=250 nm

[tex]\text{Work function of metallic caesium}=2.24\ eV\\=2.24\times 1.6021\times 10^{-19}\\=3.588704\times 10^{-19}\ Joule \ \text{(converting to SI units)}\\\lambda =\text {Wavelength of light}=250\ nm\\Energy\\E=\frac{hc}{\lambda}\\\text{Where h=Plancks constant}=6.62607004\times 10^{-34} m^2kg / s\\\text{c=speed of light}=3\times 10^8\ m/s\\\Rightarrow E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{250\times 10^{-9}}\\\Rightarrow E=0.07951284048\times 10^{-17} Joule\\[/tex]

[tex]\text{Kinetic energy}=\text{E - Work function}\\K.E.=(7.951284048\times 10^{-19})-(3.588704\times 10^{-19})\\K.E.=4.362580048\times 10^{-19}\ Joule\\[/tex]

b) When, λ=600 nm

[tex]E=\frac{hc}{\lambda}\\E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{600\times 10^{-9}}\\\Rightarrow E=3.313030502\times 10^{-19}\\\text{Kinetic energy}=\text{E - Work function}\\K.E.=(3.31303502\times 10^{-19})-(3.588704\times 10^{-19})\quad \text{Work function remains constant}\\K.E.=-0.27566898\times 10^{-19}\ Joule[/tex]

Answer:

Part B)

v = 4.98 m/s

Explanation:

Part a)

As the ball is rolling on the inclined the the friction force will be static friction and the contact point of the ball with the plane is at instantaneous rest

The point of contact is not slipping on the ground so we can say that the friction force work done would be zero.

So here in this case of pure rolling we can use the energy conservation

Part b)

By energy conservation principle we know that

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

so we will have

[tex]\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega'^2 + mgh[/tex]

here in pure rolling we know that

[tex]v = R\omega[/tex]

now from above equation we have

[tex]\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv'^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v'}{R})^2 + mgh[/tex]

now we have

[tex]\frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv'^2(1 + \frac{2}{5}) + mgh[/tex]

now plug in all values in it

[tex]\frac{1}{2}m(6^2)(\frac{7}{5}) = \frac{1}{2}mv'^2(\frac{7}{5}) + m(9.81)(0.80)[/tex]

[tex]25.2 = 0.7v^2 + 7.848[/tex]

[tex]v = 4.98 m/s[/tex]

Final answer:

The conservation of energy can be applied to the rolling ball because rolling motion involves both kinetic energy and potential energy. By equating the initial total energy of the ball to the final kinetic energy of the ball, we can find its velocity after it has passed the hill.

Explanation:

(a) The conservation of energy can be applied to the rolling ball because rolling motion involves both kinetic energy and potential energy. Although rolling requires friction, the work done by friction is accounted for in the system's total energy. Therefore, the law of conservation of energy can still be applied.

(b) To find the velocity of the ball after it has passed the hill, we can apply the conservation of energy. At the top of the hill, the ball has both potential energy and kinetic energy. At the bottom of the hill, the ball only has kinetic energy. By equating the initial total energy of the ball to the final kinetic energy of the ball, we can solve for its velocity.

Answer:

[tex]L = 9.42 kg m^2/s[/tex]

Explanation:

Angular speed of the cylinder is given as

[tex]f = 1500 rpm[/tex]

[tex]f = 1500 round/60 s[/tex]

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

[tex]\omega = 2\pi(25) = 50 \pi[/tex]

now moment of inertia of the cylinder is given as

[tex]I = \frac{1}{2}mR^2[/tex]

[tex]I = \frac{1}{2}(3)(0.2)^2[/tex]

[tex]I = 0.06 kg m^2[/tex]

now we have

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

[tex]L = (0.06)(50\pi)[/tex]

[tex]L = 9.42 kg m^2/s[/tex]

Answer:

130N

Explanation:

F=(M1+M2)V

F= (70+60)*1

F=130*1

F=130N//

Answer:

They must have been traveling at 5333.33 km/h to cover that distance in 3 days.

That speed are 6,66 times higher than the speed of an aircraft jet.

Explanation:

d= 384000 km

t= 3 days = 3*24hr = 72hr

V= 384000km/72hr

V= 5333.33 km/h

comparison:

V1/V2= 5333.33/800

V1/V2= 6.66

Answer:[tex]V_{air}=0.259m/s[/tex]

[tex]V_{water}=0.01293m/s[/tex]

Explanation:

Given data

Length of pipe[tex]\left ( L\right )[/tex]=13cm=0.13m

From tables at [tex]T=30^{\circ}[/tex]

Kinematic viscosity of air[tex]\left ( \mu\right )=1.6036\times 10{-5} m^{2}/s[/tex]

and reynolds number is given by

Re=[tex]\frac{V\times \characteristic\ length}{Kinematic visocity}[/tex]

Flow is laminar up to Re.no. 2100

Re=[tex]\frac{V\times L}{1.6036\times 10{-5}}[/tex]

2100=[tex]\frac{V\times 0.13}{1.6036\times 10{-5}}[/tex]

V=0.259 m/s

For water

Kinematic viscosity of water[tex]\left ( \mu\right )=0.801\times 10{-6}m^{2}/s[/tex]

2100=[tex]\frac{V\times 0.13}{0.801\times 10{-6}}[/tex]

V=0.01293 m/s

Explanation:

It is given that,

Frequency of monochromatic light, [tex]f=5\times 10^{14}\ Hz[/tex]

Separation between slits, [tex]d=2.2\times 10^{-5}\ m[/tex]

(a) The condition for maxima is given by :

[tex]d\ sin\theta=n\lambda[/tex]

For third maxima,

[tex]\theta=sin^{-1}(\dfrac{n\lambda}{d})[/tex]

[tex]\theta=sin^{-1}(\dfrac{n\lambda}{d})[/tex]

[tex]\theta=sin^{-1}(\dfrac{nc}{fd})[/tex]  

[tex]\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})[/tex]  

[tex]\theta=4.69^{\circ}[/tex]

(b) For second dark fringe, n = 2

[tex]d\ sin\theta=(n+1/2)\lambda[/tex]

[tex]\theta=sin^{-1}(\dfrac{5\lambda}{2d})[/tex]

[tex]\theta=sin^{-1}(\dfrac{5c}{2df})[/tex]

[tex]\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})[/tex]

[tex]\theta=3.90^{\circ}[/tex]

Hence, this is the required solution.

(a) The angle of the third bright fringe (θ₃) is approximately 4.69 degrees (b) The angle of the second dark fringe (θ₂) is approximately 3.90 degrees.

To solve this problem, we can use the formula for the angles of the maxima and minima in a Double-Slit Experiment diffraction pattern. For the third bright fringe, the condition for constructive interference is given by:

[tex]\[ d \sin(\theta_m) = m \lambda \][/tex]

where:

- d is the slit separation,

- [tex]\( \theta_m \)[/tex] is the angle of the mth bright fringe,

- m is the order of the fringe

- λ is the wavelength of the light.

For the second dark fringe, the condition for destructive interference is given by:

[tex]\[ d \sin(\theta_n) = \left( n + \frac{1}{2} \right) \lambda \][/tex]

where:

- [tex]\( \theta_n \)[/tex] is the angle of the nth dark fringe,

- n is the order of the dark fringe

Given that the frequency f is related to the wavelength λ by the speed of light (c) as c = fλ, we can express λ in terms of f.

Part (a): Third Bright Fringe

[tex]\[ d \sin(\theta_m) = m \lambda \][/tex]

[tex]\[ \sin(\theta_3) = \frac{3 \lambda}{d} \][/tex]

[tex]\[ \sin(\theta_3) = \frac{3 c}{d f} \][/tex]

[tex]\[ \theta_3 = \sin^{-1}\left(\frac{3 c}{d f}\right) \][/tex]

Part (b): Second Dark Fringe

[tex]\[ d \sin(\theta_n) = \left( n + \frac{1}{2} \right) \lambda \][/tex]

[tex]\[ \sin(\theta_2) = \frac{(2 + 0.5) \lambda}{d} \][/tex]

[tex]\[ \sin(\theta_2) = \frac{2.5 c}{d f} \][/tex]

[tex]\[ \theta_2 = \sin^{-1}\left(\frac{2.5 c}{d f}\right) \][/tex]

Now, plug in the values:

Given:

- [tex]\( d = 2.20 \times 10^{-5} \)[/tex] m,

- [tex]\( f = 5.00 \times 10^{14} \)[/tex] Hz,

- [tex]\( c = 3.00 \times 10^8 \)[/tex] m/s,

- m = 3,

- n = 2.

[tex]\[ \theta_3 = \sin^{-1}\left(\frac{3 \times 3.00 \times 10^8}{2.20 \times 10^{-5} \times 5.00 \times 10^{14}}\right) \][/tex]

[tex]\[ \theta_3 \approx 4.69^\circ \][/tex] (rounded to two decimal places, as given)

[tex]\[ \theta_2 = \sin^{-1}\left(\frac{2.5 \times 3.00 \times 10^8}{2.20 \times 10^{-5} \times 5.00 \times 10^{14}}\right) \][/tex]

[tex]\[ \theta_2 \approx 3.90^\circ \][/tex] (rounded to two decimal places, as given)

So, the correct answers are indeed [tex]\( \theta_3 \approx 4.69^\circ \)[/tex] for part (a) and [tex]\( \theta_2 \approx 3.90^\circ \)[/tex] for part (b).

Learn more about Double-Slit Experiment here:

https://brainly.com/question/28108126

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Answer:

Greatest amount of time to warm up: Aluminum

Then steel,

then copper

Explanation:

The higher the heat capacity, the longer it takes to warm the metal up.

Final answer:

To determine which metal heats up the fastest, you must consider their specific heat capacities. Copper, with the lowest specific heat capacity of 390 J/(kg°C), will heat up the fastest, followed by steel (450 J/(kg°C)) and aluminum (910 J/(kg°C)).

Explanation:

The concept of specific heat capacity is critical in understanding the rate at which different materials will heat up or cool down. Specific heat capacity, denoted by Cmetal, refers to the amount of heat needed to raise the temperature of one kilogram of a substance by one degree Celsius.

The metals in question are steel at 450 J/(kg°C), aluminum at 910 J/(kg°C) and copper at 390 J/(kg°C). The lower the specific heat capacity, the faster a material will reach a higher temperature when exposed to a constant heat source. Hence, according to their specific heat capacities, the metals rank in the following order in terms of heating up quickly: copper (390 J/(kg°C)), steel (450 J/(kg°C)), and aluminum (910 J/(kg°C)). This is because copper has the lowest specific heat capacity and therefore will require less energy to increase in temperature compared to steel and aluminum.