The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 4.80×10−19 JJ . In what direction and through what potential difference Vb−VaVb−Va does the particle move?

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

A. The angular acceleration is 5 rad/s². B. The kinetic energy of the shaft after 8 seconds is 16000 J. C. The work done by the motor in this time is 1600 J. D. The average power delivered by the motor over this time interval is 200 W.

A. Angular acceleration can be found using the rotational analog to Newton's second law a = net τ/I. The moment of inertia I is given and the torque τ can be found from the given torque value. So, τ = 10 Nm and I = 2.0 kgm2. Substituting these values, we get a = 10 Nm / 2.0 kgm2 = 5 rad/s².

B. Using the equation K = (1/2) I ω², where I is the moment of inertia and ω is the angular velocity, we can calculate the initial angular velocity ω₁ as 0 rad/s. Then, the final angular velocity ω can be found using the relationship ω = ω₁ + αt. So, ω = 0 rad/s + (5 rad/s²)(8 s) = 40 rad/s. Finally, substituting these values into the kinetic energy equation, we have K = (1/2)(2.0 kgm2)(40 rad/s)² = 16000 J.

C. The work done by the motor can be found using the equation W = τθ, where τ is the torque and θ is the angular displacement. In this case, θ = ω₁t + (1/2)αt² = 0 rad/s(8 s) + (1/2)(5 rad/s²)(8 s)² = 160 rad. Therefore, W = (10 Nm)(160 rad) = 1600 J.

D. The average power can be calculated using the equation P = W/t, where W is the work done and t is the time interval. Substituting the values we found in the previous calculations, we have P = 1600 J / 8 s = 200 W.

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Answer

given,

diameter of telescope = 2 m

time to collect area = 1 hour

Diameter of the another telescope = 6 m

The light collected by the telescope is directly proportional to the area of its primary mirror.

Area of the mirror is directly proportional to the square of diameter.

so,

6 m diameter telescope will carry (6/2)² = 9 times more light than 2 m telescope.

Time taken to collect light = 60/9 = 6.67 minutes.

now, For 12 m telescope

12 m diameter telescope will carry (12/2)² = 36 times more time than 2 m telescope.

Time taken by 12 m telescope to collect light= 60/36 = 1.7 minutes.

To find the time required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use a formula and rearrange it to solve for the time taken.

To find out how much time would be required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use the formula:

(light collected by telescope 1) × (time taken by telescope 1) = (light collected by telescope 2) × (time taken by telescope 2)

Let's solve for the time taken by the 6-m telescope and the 12-m telescope:

By rearranging the equation and solving for the time, we can find the answer.

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Answer:

(b) the point charge is moved outside the sphere

Explanation:

Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.

[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]

According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.

If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.

Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.

Answer:

854.39 N

Explanation:

The formula for the fundamental frequency of a stretched string is given as,

f = 1/2L√(T/m)..................... Equation 1

Where f = fundamental frequency, L = Length of the wire, T = Tension, m = mass per unit length.

Given: f = 261.6 Hz, L = 0.6 m, m = (5.2×10⁻³/0.6) = 8.67×10⁻³ kg/m.

Substitute into equation 1

261.6 = 1/0.6√(T/8.67×10⁻³)

Making T the subject of the equation,

T = (261.6×0.6×2)²(8.67×10⁻³)

T =854.39 N

Hence the tension of the wire is 854.39 N.

We have that for the Question "To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C "

Answer:

From the question we are told

a piano has a length of 0.6m and a mass of [tex]5.2 * 10^{−3} kg[/tex]

Generally the equation for frequency is mathematically given as

[tex]F = \frac{V}{2L}[/tex]

where,

[tex]V = \sqrt\frac{T}{U}[/tex]

Therefore,

[tex]261.6 = \frac{V}{2*0.76}\\\\V = 261.6*2*0.6\\\\V = 313.92m/s[/tex]

so,

[tex]v^2 = \frac{T}{U}\\\\T = V^2 * U\\\\T = 512.43N[/tex]

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Answer : The volume of the second tank is, [tex]4ft^3[/tex]

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 2 atm

[tex]P_2[/tex] = final pressure of gas = 1 atm

[tex]V_1[/tex] = initial volume of gas = [tex]2ft^3[/tex]

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get:

[tex]2atm\times 2ft^3=1atm\times V_2[/tex]

[tex]V_2=4ft^3[/tex]

Therefore, the volume of the second tank is, [tex]4ft^3[/tex]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

[tex]F_k = F_{W,E}[/tex]

[tex]kx_1 = mg[/tex]

The extension of the spring due to the weight of the object on Moon is a value of [tex]x_2[/tex], then

[tex]kx_2 = mg_m[/tex]

Recall that gravity on the moon is a sixth of Earth's gravity.

[tex]kx_2 = m\frac{g}{6}[/tex]

[tex]kx_2 = \frac{1}{6} mg[/tex]

[tex]kx_2 = \frac{1}{6} kx_1[/tex]

[tex]x_2 = \frac{1}{6} x_1[/tex]

We have that the displacement at the earth was [tex]x_1 = 0.3m[/tex], then

[tex]x_2 = \frac{1}{6} 0.3[/tex]

[tex]x_2 = 0.05m[/tex]

Therefore the displacement of the mass on the spring on Moon is 0.05m

The same spring and mass system will have a displacement of 0.05 m on the Moon compared to 0.3 m on Earth due to the Moon's lower gravitational acceleration.

The subject of the question relates to how a spring and mass system will behave on Earth versus the Moon. The solution requires understanding of Hooke's Law and gravitation. Hooke's Law states that the displacement of a spring is directly proportional to the force applied. On Earth, when a 0.500 kg mass is suspended and displaced by 0.3 m, this sets up a certain relationship of force (F = mg, where g is Earth's gravity, 9.8 m/s²): F = 0.500 kg * 9.8 m/s² = 4.9 Newtons. Displacement, d, is proportional to the force:F = kd, so d = F/k where k is the spring constant.

On the Moon, g is not equal to Earth's, it's 1.67 m/s². The same hanging mass on the Moon would exert a force of Fm=0.500 kg * 1.67 m/s² = 0.835 Newtons. Since the spring constant doesn't change, and F = kd still holds, the new displacement on the Moon will be greater because the force is smaller. Displacement on the Moon (dM) will be dM = Fm/k, which is Fm divided by the same k we had on Earth. Not knowing k, we do know dM = (Fm/F) * d, and (Fm/F) is equal to the ratio of the Moon's gravity to Earth's gravity, 1/6, so dM = (1/6)*0.3 m = 0.05 m.

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Explanation:

The visible light coming from celestial bodies from space are affected most by the atmosphere. The visible light suffer blurring and absorption.  So, to avoid such situation optical telescopes are kept in mountains where atmosphere is thin.

Whereas radio signal are not affected by the atmosphere so, they need not be placed in mountains.However, it is affected by various noises from the man made devices. So, to avoid these noises radio telescopes are kept in deep valleys.

Final answer:

Optical telescopes are placed on mountains to minimize atmospheric interference, reduce light pollution, and capture clearer images, while radio telescopes are located in valleys to shield them from man-made radio interference for clearer radio signal observations.

Explanation:

Optical astronomers often put their telescopes at the tops of mountains because these locations offer several advantages that are critical for observing the cosmos. Sir Isaac Newton mentioned that a serene and quiet air, often found on the tops of high mountains, is beneficial for reducing the confusion of rays caused by the atmosphere's tremors. This is echoed by modern practices where observatories are located in high, dry, and dark sites to minimize atmospheric interference, reduce light pollution, and avoid water vapor which absorbs infrared light.

The Andes Mountains in Chile, the desert peaks of Arizona, and the summit of Maunakea in Hawaii are examples of such ideal locations. On the other hand, radio astronomers sometimes place their telescopes in deep valleys to shield them from radio interference from human-made sources such as cell phones and electrical circuits. The natural geography of valleys can act as a barrier against these disturbances, providing a clearer signal for radio observations.

Answer:

0.00000103149529075 m²/s

0.0103149529075 stokes

Explanation:

A = [tex]2.4\times 10^-5[/tex]

B = 250 K

C = 140 K

T = 20°C

[tex]\rho[/tex] = Density of water = 998 kg/m³

Viscosity is given by

[tex]\mu=A\times 10^{\dfrac{B}{T-C}}\\\Rightarrow \mu=2.4\times 10^{-5}\times 10^{\dfrac{250}{273.15+20-140}}\\\Rightarrow \mu=0.00102943230017\ Pas[/tex]

Kinematic viscosity is given by

[tex]\nu=\dfrac{\mu}{\rho}\\\Rightarrow \nu=\dfrac{0.00102943230017}{998}\\\Rightarrow \nu=0.00000103149529075\ m^2/s[/tex]

The kinematic viscosity is 0.00000103149529075 m²/s

In BG units [tex]0.00000103149529075\times 10^4=0.0103149529075\ stokes[/tex]

In the SI system of units, the dynamic viscosity of water can be computed using the equation μ=A10B/(T-C). The dimensions of A are (m⋅s²)/kg. To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The kinematic viscosity of water at 20°C is approximately 0.0001 m²/s in SI units and 1 cm²/s in BG units.

(a) Dimensions of A:

To determine the dimensions of A, we need to rearrange the equation and analyze the units on both sides. We know that viscosity has units of kilograms per meter per second (kg/m/s). Plugging in the given values for B and C, we have μ=A10B/(T-C). The dimensions of 10B/(T-C) are (dimensionless). Therefore, the dimensions of A must be kg/m/s multiplied by the inverse of the dimensions of 10B/(T-C), which is 1/(kg/m/s) or (m⋅s²)/kg.

(b) Kinematic viscosity at 20°C:

To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The density of water at 20°C is approximately 998 kg/m³. Plugging in the given values for A and C, we can calculate μ using the equation μ=A10B/(T-C). Substituting the values into the formula ν=μ/ρ, we get ν≈μ/ρ≈(A10B/(T-C))/ρ.

In the SI unit, ν≈(2.4×10-5×10250 K/(293 K-140 K))/998 kg/m³≈0.0001 m²/s.

In the BG unit, ν≈0.0001 m²/s×104 cm²/1 m²≈1 cm²/s.

Answer:

0.08 N/C

Explanation:

Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,

E = Kq/r².............................. Equation 1

Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.

making q the subject of the equation,

q = Er²/k............................... Equation 2

Given: E = 2 N/C, r = 4 m,

Substitute into equation 2

q = 2(4)²/k

q = 32/k C.

When r is increased to 20 m,

E = k(32/k)/20²

E = 32/400

E = 0.08 N/C.

Hence the electric Field = 0.08 N/C

The initial electric field from a charge is 2 N/C at 4m distance. When the distance increases fivefold, the strength of the electric field decreases by the square of this factor, resulting in a new electric field magnitude of 0.08 N/C.

This question is about the relationship between an electric field and its distance from a point charge. According to the formula for the magnitude of the electric field generated by a point charge, which is E = kQ / r^2, where E is the electric field strength, k is Coulomb's constant, Q is the charge, and r is the distance from the charge, the electric field is inversely proportional to the square of the distance.

At a distance of 4m, the electric field's magnitude is 2 N/C. If you increase the distance to 20m (which is 5 times more than the initial distance), the new electric field magnitude will be the initial magnitude divided by the square of this factor, i.e., 2 N/C divided by 5^2 = 2 N/C / 25, which equals to 0.08 N/C.

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Answer:

v = 0.2035 m/s

Explanation:

given,

Amplitude,A = 0.0127 m

Frequency, f = 2.55 Hz

maximum speed of the needle = ?

we know,

x = A cos ω t

velocity of the motion

v = - A ω sin ω t

for maximum speed

sin ω t = 1

v = - A ω

ω = 2 π f

now,

v =  A 2 π f

v = 0.0127 x 2 π x 2.55

v = 0.2035 m/s

Hence, the maximum speed of the needle is equal to 0.2035 m/s

Answer:

Explanation:

Each square meter of earth surface intercepts 1110 protons per second.

Total surface of the earth will intercept  no of proton equal to

= 1110 x area of surface

= 1110 x 5.1 x 10¹⁴ m²

= 5661 x 10¹⁴

Total charge

= 1.6 x 10⁻¹⁹ x  5661 x 10¹⁴ coulomb per second

= .09 A .

electric current intercepted by the total surface area of the planet = .09 A.

Explanation:

Exothermic reaction is defined as the reaction in which release of heat takes place. This also means that in an exothermic reaction, bond energies of reactants is less than the bond energies of products.

Hence, difference between the energies between the reactants and products releases as heat and therefore, enthalpy of the system will decrease.

Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.

As water acts as a coolent and when fuel rods in a nuclear reactor are immersed in it then heat created by coolent is absorbed by water and then it changes into steam.

Since, absorption of heat occurs in the nuclear reactor. Therefore, it is an endothermic reaction.

Thus, we can conclude that nuclear reactors use fuel rods to heat water and generate steam. This process is endothermic.

Answer:

The heat produced by the nuclear reactor is an exothermic process while the heat absorbed by the water to convert into steam is an endothermic process.

Explanation:

Nuclear reactor being the heat of a nuclear power plant uses the radioactive uranium fuel to generate the heat by the process of nuclear fission in a controlled manner.

The processing of uranium is carried out into small ceramic pellets which are stacked together into sealed metal tubes known as fuel rods.

Usually more than 200 such rods are bunched together leading to the formation of a fuel assembly.

The core of the reactor is often made up of a couple hundred assemblies, according to its power level.

Inside the reactor vessel, these fuel rods are immersed into water which serve as both a coolant and moderator. The moderator helps slow down the neutrons produced by fission to sustain the chain reaction.

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

Answer:

[tex]L=4.19*10^{-6}H[/tex]

Explanation:

The self-inductance of a solenoid is defined as:

[tex]L=\frac{\mu N^2A}{l}[/tex]

Here [tex]\mu_0[/tex] is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:

[tex]L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H[/tex]

The self-inductance of the given solenoid would be approximately 1.395x10^-5 H. The calculation involved substituting known variables into the formula for self-inductance and solving.

The self-inductance of a solenoid can be calculated using the formula: L = µ₀n²A/l where L is the self-inductance, µ₀ is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area, and l is the length of the solenoid.

In this scenario, the cross-sectional area (A) = 1.00 × 10⁻⁴ m², length of the solenoid (l) = 30.0 cm = 0.3m, and the number of turns (n) = 100, so n (number of turns per unit length) = 100/0.3 = 333.33 turns/m. µ₀, the permeability of free space is a constant which is given as 4π × 10⁻⁷ T m/A. So, substituting these values into the formula will give:

L (self-inductance) = 4π × 10⁻⁷ T m/A * (333.33 /m)² * 1.00 × 10⁻⁴ m² / 0.3 m = 1.395x10^-5 H, (where H indicates henries, the unit for inductance)

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Answer:

Because of Moment of inertia.

Explanation:

Larger wheels have larger moment of inertia,when they rotate rotational energy is stored in spinnning of motion so they rotate slowly while lighter wheels move faster comparitively because they use smaller moment of inertia.The thing is to keeping tyre in contact with road,when a vehicla hits a

jolt it is difficult with heavy tyre to be in contact with road. When vehicle loses contact the driver will lose steering control which result in giving a way to sliding friction.

Answer:

Explanation:

Given

First capacitor magnitude [tex]C=1.75\ \mu F[/tex]

Second capacitor magnitude [tex]C=6.00\ \mu F[/tex]

Voltage of battery [tex]V=3.00\ V[/tex]

Both capacitor are connected in series so net capacitor is given by

[tex]\frac{1}{C_{net}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]

[tex]\frac{1}{C_{net}}=\frac{1}{1.75}+\frac{1}{6}[/tex]

[tex]C_{net}=\frac{6\times 1.75}{1.75+6}[/tex]

[tex]C_{net}=1.35\ \mu F[/tex]

So charge Across each capacitor is given by

[tex]Q=C_{net}\times V[/tex]

[tex]Q=1.35\times 10^{-6}\times 3[/tex]

[tex]Q=4.064\ \mu C[/tex]

Answer:

The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

The excess charge on the surface of the earth is 4.55×10⁵C

To calculate the charge on the surface of the earth, we apply the equation of electric field strength, that is:

[tex]E = k\frac{Q}{r^2}[/tex]

here, E is the electric field strength = 100N/C (given)

k = 9×10⁹ Nm²/C²

Q is the charge, and

r = distance = radius of earth = 6.4×10⁶ m

Now,

[tex]Q = \frac{r^2E}{k}\\ \\=\frac{(6.4*10^6)^2*100}{9*10^9}\\\\Q=4.55*10^5 C[/tex] is the charge acquired on the surface of the earth.

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To solve this problem we will apply the concepts related to electrostatic energy, defined as,

[tex]U_e = \frac{kq_1q_2}{d^2}[/tex]

Here,

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object (electron and proton, at this case the same)=

d = Distance

Replacing with our values we have that

[tex]U_e = \frac{(9*10^9Nm^2)(1.6*10^{-19}C)(-1.6*10^{-19}C)}{0.0539*10^{-9}m}[/tex]

[tex]U_e = -4.27*10^{-18}J[/tex]

Therefore for the Part A the answer is [tex]-4.27*10^{-18}J[/tex] and por the Part B the sign indicates that the force between the proton and electron is attractive

Answer:

103.3°C

Explanation:

k = Thermal conductivity of stainless steel = 14 W/m² K

P = Power = 800 W

d = Diameter = 22 cm

t = Thickness of kettle = 2.2 mm

[tex]\Delta T[/tex] = Change in temperature = [tex](T_2-100)[/tex]

100°C boiling point of water

Power is given by

[tex]P=\dfrac{kA\Delta T}{t}\\\Rightarrow P=\dfrac{kA(T_2-T_1)}{t}\\\Rightarrow T_2=\dfrac{Pt}{kA}+T_1\\\Rightarrow T_2=\dfrac{800\times 2.2\times 10^{-3}}{14\times \dfrac{\pi}{4} 0.22^2}+100\\\Rightarrow T_2=103.3\ ^{\circ}C[/tex]

The temperature of the bottom of the surface 103.3°C

The temperature of the bottom surface of the kettle is approximately 1.593 Kelvin higher than the boiling water.

To calculate the temperature of the bottom surface of the kettle, we can use the formula for heat conduction: q = (k * A * ΔT) / t. Here, q is the heat transfer rate, k is the thermal conductivity of stainless steel, A is the surface area of the bottom of the kettle, ΔT is the temperature difference between the bottom surface of the kettle and the boiling water, and t is the thickness of the bottom of the kettle. Rearranging the formula, we can solve for ΔT as:

ΔT = (q * t) / (k * A)

Plugging in the given values, we find:

ΔT = (800 W * 0.0022 m) / (14 W/(m·K) * π * (0.22 m/2)²)

Calculating this, we find ΔT ≈ 1.593 K. Therefore, the temperature of the bottom surface of the kettle would be approximately 1.593 Kelvin higher than the boiling water.

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Answer:

(a). The mass of the block of ice is 62.8 kg.

(b). The distance is 24.192 m.

Explanation:

Given that,

Horizontal force = 80.5 N

Distance = 13.0 m

Time = 4.50 s

(a). We need to calculate the acceleration of the block of ice

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]13=0+\dfrac{1}{2}\times a\times(4.50)^2[/tex]

[tex]a=\dfrac{13\times2}{(4.50)^2}[/tex]

[tex]a=1.28\ m/s^2[/tex]

We need to calculate the mass of the block of ice

Using formula of force

[tex]F = ma[/tex]

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

Put the value into the formula

[tex]m=\dfrac{80.5}{1.28}[/tex]

[tex]m=62.8\ kg[/tex]

(b). If the worker stops pushing at the end of 4.50 s,

We need to calculate the velocity

Using equation of motion

[tex]v =u+at[/tex]

Put the value into the formula

[tex]v=0+1.28\times4.50[/tex]

[tex]v=5.76\ m/s[/tex]

We need to calculate the distance

Using formula of distance

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

[tex]d=v\times t[/tex]

Put the value into the formula

[tex]d=5.76\times4.20[/tex]

[tex]d=24.192\ m[/tex]

Hence, (a). The mass of the block of ice is 62.8 kg.

(b). The distance is 24.192 m.

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

[tex]A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\[/tex]

a) The Potential difference across a 2.00 in length of a 14-gauge copper  

   wire:

  L= 2.00 m

From Table  Copper Resistivity [tex]p[/tex]= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

[tex]R=\frac{pL}{A}[/tex]

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

[tex]R=\frac{pL}{A}[/tex]

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

Answer:

θ₂ = 35.26°

Explanation:

given,

refractive index of air, n₁ = 1

refractive index of glass, n₂ = 1.5

angle of incidence, θ₁ = 60°

angle of refracted light, θ₂ = ?

using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

1 x sin 60° = 1.5 sin θ₂

sin θ₂ = 0.577

θ₂ = sin⁻¹(0.577)

θ₂ = 35.26°

Hence, the refracted light is equal to  θ₂ = 35.26°

When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The angle of the refracted light beam from the normal in air is 90 degrees.

When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The formula for Snell's law is:

n1×sin(θ1) = n2×sin(θ2)

In this case, the light beam is passing from glass (with a refractive index of 1.5) to air. Given that the angle of incidence is 60 degrees, we can use Snell's law to find the angle of refraction from the normal in air.

Using Snell's law, we have:

1.5×sin(60) = 1×sin(θ2)

Simplifying the equation, we find:

sin(θ2) = 1.5×sin(60) = 1.5×0.87 = 1.31

However, the maximum value for the sine function is 1, so the angle of refraction cannot exceed 90 degrees (the maximum angle for a light beam in air). Therefore, the angle of the refracted light beam from the normal in air is 90 degrees.

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Answer:

616.223684211 N

Explanation:

[tex]F_r[/tex] = Resistive force on the wheel = 115 N

F = Force acting on sprocket

[tex]r_2[/tex] = Radius of sprocket = 4.75 cm

[tex]r_1[/tex] = Radius of wheel = 25 cm

Moment of inertia is given by

[tex]I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2[/tex]

Torque

[tex]\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm[/tex]

Torque is given by

[tex]\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N[/tex]

The force on the chain is 616.223684211 N

Answer:

451977.40113 N/C

Explanation:

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

[tex]\sigma[/tex] = Surface charge density = [tex]4\ pc/mm^2[/tex]

Electric field near the surface of a charged conductor is given by

[tex]E=\dfrac{\sigma}{\epsilon_0}\\\Rightarrow E=\dfrac{4\times 10^{-12}\times 10^{6}}{8.85\times 10^{-12}}\\\Rightarrow E=451977.40113\ N/C[/tex]

The electric field is 451977.40113 N/C

Incomplete question.The complete question is here

If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit? The light wavelength is 500 nm and the slit width is 0.16 mm.

Answer:

The distance between first and second minimum is 1.92m

Explanation:

Given data

Wavelength of light λ=500nm

Slit width D=0.16 mm

The distance between first and second minima is 6.0 mm

To find

Distance L between the screen and the slit

Solution

As we know that

Yn=(nλL)/D

Where is L is distance between the slit and screen

D is slit width

n is order of minimum

Yn is distance of nth minimum from center maximum

λ is wavelength of light

The Distance between first and second minimum is given as:

Y₂-Y₁=6 mm

(2λL)/D-(1λL)/D=6 mm

(2λL-1λL)/D=6 mm

(λL/D)=6 mm

[tex]L=\frac{6mm(0.16mm)}{500nm}\\ L=1.92m[/tex]

The distance between first and second minimum is 1.92m

Final answer:

The screen distance is 0.19 meters, calculated using the separation between minima, slit width, and light wavelength in single-slit diffraction.

Explanation:

Solve for the distance between the screen and the slit (L).

1. Given Values:

Separation between minima (Δx) = 6.0 mm (convert to meters: 0.0060 m)

Wavelength of light (λ) = 500 nm (convert to meters: 500 x 10⁻⁹ m)

Slit width (a) = 0.16 mm (convert to meters: 0.16 x 10⁻³ m)

2. Relate Separation and Screen Distance:

We can use the relationship between the angular separation (Δθ) and the linear separation (Δx) on the screen:

Δθ ≈ Δx / L

3. Relate Separation and Minimum Position:

We know the separation (Δθ) is related to the difference between the positions of the first (m = 1) and second (m = 2) minima:

Δθ = θ₂ - θ₁ = (λ / a)

4. Combine Equations:

Substitute the equation for Δθ from step 3 into the equation from step 2:

(λ / a) ≈ Δx / L

5. Solve for Screen Distance (L):

Now we can arrange the equation to solve for L:

L ≈ Δx * a / λ

6. Plug in the Values:

L ≈ (0.0060 m) * (0.16 x 10⁻³ m) / (500 x 10⁻⁹ m)

7. Calculate and Round:

L ≈ 0.192 m (round to two significant figures)

Therefore, the distance between the screen and the slit is approximately 0.19 meters.

Answer:

v = 7.5*10⁶ m/s

Explanation:

While accelerating through a potential difference of 160 V, the electron undergoes a change in the electric potential energy, as follows:

ΔUe = q*ΔV = (-e)*ΔV = (-1.6*10⁻¹⁹ C) * 160 V = -2.56*10⁻¹⁷ J (1)

Due to the principle of conservation of energy, in absence  of non-conservative forces, this change in potential energy must be equal to the change in kinetic energy, ΔK:

ΔK = Kf -K₀

As the electron accelerates from rest, K₀ =0.

⇒ΔK =Kf = [tex]\frac{1}{2}*me*vf^{2}[/tex] (2)

From (1) and (2):

ΔK = -ΔUe = 2.56*10⁻¹⁷ J = [tex]\frac{1}{2}*me*vf^{2}[/tex]

where me = mass of the electron = 9.1*10⁻³¹ kg.

Solving for vf:

[tex]vf =\sqrt{\frac{2*(2.56e-17J)}{9.1e-31kg} } =7.5e6 m/s[/tex]

⇒ vf = 7.5*10⁶ m/s

Final answer:

The speed of an electron after accelerating from rest through a potential difference of 160 V can be calculated using the conservation of energy principle, resulting in a velocity of approximately 5.93 × 10^6 m/s.

Explanation:

To calculate the speed of an electron after it accelerates from rest through a potential difference, we use the concept of conservation of energy where the electrical potential energy converted into kinetic energy is expressed as qAV = ½mv². Here, q is the charge of the electron (-1.60 × 10^-19 C), V is the potential difference (160 V), m is the mass of the electron (9.11 × 10^-13 kg), and v is the velocity of the electron we need to find. Rearranging the equation to solve for v and substituting the given values gives:

v = sqrt(2 × q × V / m) = √(2 × (-1.60 × 10^-19 C) × 160 V / (9.11 × 10^-13kg))

Performing the calculation yields a velocity of approximately 5.93 × 10^6 m/s.

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

[tex]cos(\alpha) = 80/320 = 0.25[/tex]

[tex]\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0[/tex] relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

[tex]320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h[/tex]

Answer:

a) Ф=14°, north of west

b) 310 km/h

Explanation:

we have,

[tex]v_{p/G}[/tex], velocity of plane relative to the ground(west)

[tex]v_{p/A}[/tex],velocity of plane relative to the air(320 km/h)

[tex]v_{A/G}[/tex],velocity of air relative to the ground(80 km/h, due south).

                 [tex]v_{p/G}[/tex]=[tex]v_{p/A}[/tex]+[tex]v_{A/G}[/tex].........(1)

a)     sin(Ф)=[tex]\frac{v_{A/G}}{v_{p/A}}[/tex]

                 =[tex]\frac{80km/h}{320km/h}[/tex]              

Ф=14°, north of west

b)    using Pythagorean theorem

[tex]v_{p/G}=\sqrt{v^2_{p/A}+v^2_{A/G}}[/tex]

[tex]v_{p/G}[/tex]=310 km/h

note:

diagram is attached

Answer:

[tex]d=691.71km[/tex]

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

[tex]t=\frac{d}{v_t}-\frac{d}{v_l}[/tex]

Here d is the distance at which the earthquake take place and [tex]v_t, v_l[/tex] is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

[tex]t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km[/tex]

Answer:

Range will become 4 times of initial range

Explanation:

Let the velocity of projection is u

And angle at which projectile is projected is [tex]\Theta[/tex]

And acceleration due to gravity is [tex]g\ m/sec^2[/tex]

So range of projectile is equal to [tex]R=\frac{u^2sin2\Theta }{g}[/tex]........eqn 1

Now in second case it is given that velocity of launching is doubled

So new velocity [tex]u_{new}=2u[/tex]

So new range will be equal to [tex]R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}[/tex] .....eqn 2

Now dividing eqn 2 by eqn 1

[tex]\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }[/tex]

[tex]R_{new}=4R[/tex]

So if we double the initial launch speed then range will become 4 times

Doubling the initial launch speed while keeping the launch angle fixed quadruples the range of a projectile due to the direct proportionality between the range and the square of the initial speed, considering factors like launch angle, gravity, and air resistance.

If you keep the launch angle fixed, but double the initial launch speed, the range of a projectile will increase. When the initial speed is doubled, the range increases four times because the range is directly proportional to the square of the initial speed. This phenomenon is influenced by factors such as launch angle, gravity, and air resistance.

Answer:

(A) When you hold the ball still in your hands after catching it

(E) When you hold the ball still in your hands before it is thrown.

Explanation:

According to Newton's 1st law, objects subjected to 0 net force would maintain a constant velocity or staying at rest. This is not the case for the ball when it leaves your hands. This ball would always be subjected to gravitational acceleration g so its velocity changes. So only (A) and (E) are correct when the ball stays still in your hands.

Final answer:

A zero net force acts on the basketball when it is held still after catching it (A) and before throwing it (E) because these are the points where there is no acceleration due to balanced forces. Points B, C, and D have non-zero net forces because the ball is accelerating due to the force of gravity.

Explanation:

When considering a basketball being thrown straight up and ignoring air resistance, a zero net force acts on the basketball:

(A) When holding the ball still after catching it because there is no acceleration and gravity is balanced by the upward force from your hands.

(E) When holding the ball still before throwing it for the same reason as in (A).

The points at which the net force is not zero are:

(B) Just after the ball leaves the hand as the only force acting on the ball is gravity, resulting in acceleration and therefore a net force downwards.

(C) At the highest point because gravity is still acting downwards, although the ball is temporarily at rest.

(D) As the ball hits the hands because the hands apply an upward force to decelerate the ball, which is different from the force of gravity, hence not zero net force.

Overall, the net force on an object is related to its acceleration due to Newton's second law of motion, and since acceleration occurs whenever the ball is in motion and not at rest, there is a net force acting on the ball during these periods.