A nonconducting sphere contains positive charge distributed uniformly throughout its volume. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity. (There may be more than one correct choice)

a) The potential is highest at the center of the sphere. b) The potential at the center of the sphere is the same as the potential at the surface. c) The potential at the surface is higher than the potential at the center. d) The potential at the center is the same as the potential at infinity. e) The potential at the center of the sphere is zero.

Answer:

5.062 x 10^-9 mg

Explanation:

Charge on glass rod = 89 uC

Charge on silk cloth = - 89 uC

mass of scarf, M = 100 g

Number of excess electrons on silk cloth = charge / charge of one electron

n = (89 x 10^-6) / ( 1.6 x 10^-19)

n = 5.5625 x 10^14

mass of one electron = 9.1 x 10^-31 kg

Mass of  5.5625 x 10^14 electrons =  5.5625 x 10^14 x 9.1 x 10^-31 kg

ΔM = 5.062 x 10^-16 kg

Δ M / M = (5.062 x 10^-16) / 0.1 = 5.062 x 10^-15 kg = 5.062 x 10^-9 mg

It is not noticeable.

Answer:

Intensity of a wave is Power per unit area.

Explanation:

The intensity of wave is defined as the power per unit area in the direction of motion of wave i.e.

[tex]I=\dfrac{P}{A}[/tex]

Also, the intensity of a wave is directly proportional to the square of its amplitude. Let a is the amplitude of a wave. So,

[tex]I\propto a^2[/tex]

The SI unit of power is watt and the SI unit of area is m². So, the SI unit of intensity is W/m². Hence, the correct option is (a) "Intensity of a wave is Power per unit area".

Answer:

980 kJ

Explanation:

Work = change in energy

W = mgh

W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)

W = 980,000 J

W = 980 kJ

The pump does 980 kJ of work.

The work done by the pump against gravity in transfering 5 m³ of water to a tank 20 m above the lake is 980000 J

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 5 m³

Density of water = 1000 Kg/m³

Mass = Density × Volume

Mass of water = 1000 × 5

Mass of water (m) = 5000 Kg

Height (h) = 20 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done = mgh

Work done = 5000 × 9.8 × 20

Thus, the work done by the pump is 980000 J

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The question requires determining angles for bright and dark fringes in a double-slit experiment using the formula d sin(\theta) = m\lambda. The student should apply this formula, adjusting it for dark fringes, and solve for the angles for each order m, considering the wavelength and slit separation given.

The question relates to Young's double-slit experiment, which demonstrates the wave nature of light through the interference pattern produced when light passes through two closely spaced slits. The angles locating the fringes can be calculated using the formula

d sin(\theta) = m\lambda, where d is the separation between the slits, \theta is the angle of the fringe from the central maximum, m is the order of the fringe (which can be an integer or half-integer value depending on whether it's a bright or dark fringe), and \lambda is the wavelength of the light.

(a) For the dark fringe with m=0, we expect no fringe to appear as m=0 corresponds to the central maximum which is bright, not dark.

(b) For the bright fringe with m=1, we rearrange the formula to \theta = arcsin(m\lambda / d). Substituting the given values, the angle can be calculated.

(c) For the dark fringe with m=1, the condition is changed to d sin(\theta) = (m + 0.5)\lambda, as dark fringes occur at half-wavelength shifts from the bright fringes.

(d) The calculation for the bright fringe with m=2 follows the same procedure as for m=1, using the appropriate value for m.

Answer:

Vertically upwards

Explanation:

Fleming's left hand rule: it states taht when we spread our fore finger, middle finger and thumb in such a way that they are mutually perpendicular to each other, theN thumb indicates the direction of force, fore finger indicates the direction of magnetic field and the middle finger indicates the direction of current.

So according to this rule the direction of force is vertically upwards.

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

[tex]v_i = 0[/tex]

also the acceleration of the stone is due to gravity which is given as

[tex]g = 32 ft/s^2[/tex]

now we have

[tex]v_f = 180 ft/s[/tex]

so from above equation

[tex]180^2 - 0 = 2(32)d[/tex]

[tex]d = 506.25 ft[/tex]

Using the equation of motion, we find that the height of the roof from which the stone was dropped is approximately 506.25 feet.

The subject question involves a physics concept relating to the motion of objects under the influence of gravity, in particular, the calculation of displacement during free fall. We can solve this problem using one of the fundamental equations of motion: final velocity2 = initial velocity2 + 2*a*d. In this case, initial velocity (u) = 0 (as the stone was dropped), final velocity (v) = -180 feet per second (negative as the stone is moving downwards), and acceleration (a) due to gravity is -32 feet/second2 (negative as gravity acts downward).

Substituting these values, we get:
(-180)2 = (0)2 + 2*(-32)*d.

This simplifies to:
32400 = -64d.

Solving for d (which represents the height of the roof), we get:
d = -32400 / -64 = 506.25 feet. Hence, the roof is approximately 506.25 feet high.

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R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

To determine the resistance of this wire is equal to 0.4082 Ohms.

Given the following data:

To determine the resistance of this wire:

First of all, we would determine the area of the wire by using this formula:

[tex]A =\frac{\pi d^2}{4} \\\\A =\frac{3.142 \times (5\times 10^{-4})^2}{4}\\\\A = 1.96 \times 10^{-7}\;m^2[/tex]

Now, we can determine the resistance of this wire:

[tex]R=\frac{\rho L}{A} \\\\R=\frac{3.2 \times 10^{-8} \times 2.5}{1.96 \times 10^{-7}}[/tex]

Resistance = 0.4082 Ohms.

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Answer:

[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]

Explanation:

Here the given values corresponds to the density of a substance

The density of a substance in standard SI unit is given as Kg/m³.

Given value = 43.28lbf/in³

also

we know,

1 lbf = 0.45kg

1in = 0.0254 m

thus,

[tex]\frac{1lbf}{1in^3}=\frac{0.45kg}{0.0254^3m^3}=27460.68\frac{kg}{m^3}[/tex]

therefore,

[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]

Answer:

6.03 x 10^-3 C/Kg

Explanation:

E = 3.8 x 10^4 N/C, u = 2.32 m/s, s = 5.98 cm = 0.0598 m, t = 0.2 s, g = 9.8 m/s^2

Acceleration on object is a .

Use second equation of  motion.

S = u t + 1/2 a t^2

0.0598 = 2.32 x 0.2 + 0.5 x a x 0.2 x 0.2

0.0598 = 4.64 + 0.02 x a

a = - 229 m/s^2

Now, F = ma = qE

q / m = a / E = 229 / (3.8 x 10000)

q / m = 6.03 x 10^-3 C/Kg

Calculate the charge-to-mass ratio of the object given initial speed, distance, time, and electric field strength.

Given: Electric field magnitude = 3.80 x 10^4 N/C, initial speed = 2.32 m/s, distance traveled = 5.98 cm, time = 0.200 s, acceleration due to gravity = 9.80 m/s^2.

To find: The object's charge-to-mass ratio (q/m).

Calculations: Determine the acceleration using the provided data, then relate the forces acting on the object (gravity and electric field) to find q/m.

Answer:

[tex]v_a = 2 v_b[/tex]

Explanation:

As we know that the speed of car A and car B is given by

[tex]v_a[/tex] & [tex]v_b[/tex]

now we know that both cars are decelerated by same deceleration and stopped finally

so the distance moved by the car is given by the equations

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex]0 - v_a^2 = 2(-a) d_a[/tex]

[tex]d_a = \frac{v_a^2}{2a}[/tex]

similarly we have

[tex]d_b = \frac{v_b^2}{2a}[/tex]

now we know that

[tex]d_a = 4 d_b[/tex]

[tex]\frac{v_a^2}{2a} = 4 \frac{v_b^2}{2a}[/tex]

[tex]v_a = 2 v_b[/tex]

Answer:

Motorcycle travel 39.75 m while coming to a stop

Explanation:

We have equation of motion

        v = u + at

   Final velocity, v = 0 m/s, u = 15 m/s, t = 5.3 s

        0 = 15 + a x 5.3

         a = -2.83 m/s²

 Now we have the other equation of motion

        v² = u² + 2as

        0² = 15² - 2 x 2.83 x s

         s = 39.75 m

Motorcycle travel 39.75 m while coming to a stop

Ham como assim pq ta em ingles

Final answer:

To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media.

Explanation:

To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media:

n1 * sin(θ1) = n2 * sin(θ2)

Given that the incident angle is 50.0° and the refracted angle is 27.7°, we can plug in the values to find the index of refraction:

n1 * sin(50.0°) = n2 * sin(27.7°)

The index of refraction of the glass is the ratio of the index of refraction of the glass to the index of refraction of air, which is approximately 1:

n2 / n1 = 1 / sin(27.7°)

Answer:

option c) 0.88c is the correct answer

Explanation:

using the Lorrentz equation we have

[tex]t=\frac{d}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]

where,

t = time taken to cover the distance

d = Distance

v = velocity

c = speed of light

given

d = 15 light years

Now,

[tex]8years=\frac{15years\times c}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]

or

[tex](\frac{v}{c})^2\times (\frac{8}{15})^2=1-(\frac{v}{c})^2[/tex]

or

[tex](\frac{v}{c})^2\times 0.284=1-(\frac{v}{c})^2[/tex]

or

[tex](\frac{v}{c})^2\times 0.284 + (\frac{v}{c})^2 =1[/tex]

or

[tex](1.284\times \frac{v}{c})^2 =1[/tex]

or

[tex] (\frac{v}{c}) =\sqrt{\frac{1}{0.284}}[/tex]

or

[tex]v=0.88c[/tex]

Answer:

25.2 cm

Explanation:

K = 1050 N/m

m = 1.95 kg

h = 1.75 m

By the conservation of energy, the potential energy of the block is converted into the potential energy stored in the spring

m g h = 1/2 x k x y^2

Where, y be the distance by which the spring is compressed.

1.95 x 9.8 x 1.75 = 1/2 x 1050 x y^2

33.44 = 525 x y^2

y = 0.252 m

y = 25.2 cm

The spring will compress by approximately 25.3 cm when a 1.95 kg block is dropped from a height of 1.75 m due to the conversion of gravitational potential energy into elastic potential energy.

The student's question is regarding the compression of a spring due to the gravitational potential energy of a mass that was dropped onto it. This problem can be solved by using the conservation of energy principle, where the potential energy of the block (due to its height) is converted into the spring's elastic potential energy when the spring is compressed.

To find the compression distance (x), we use the following steps:

First we calculate the gravitational potential energy:

PE = mgh = (1.95 kg)(9.81 m/s2)(1.75 m) = 33.637875 J

Then we set this equal to the spring's potential energy and solve for x:

33.637875 J = 1/2 (1050 N/m) x2

x2 = (2 * 33.637875 J) / (1050 N/m)

x2 = 0.064067619

x = 0.253 J

Therefore, the spring compresses by approximately 0.253 meters or 25.3 centimeters.

Final answer:

The question revolves around the final temperature of a mixture of ice and water in a thermally insulated system. The specific heats and enthalpy of fusion are required to solve, but the proper method involves using the conservation of energy.

Explanation:

The student is asking about the final temperature of a system consisting of a 0.0811 kg ice cube at 0°C and 0.397 kg of water at 14.8°C. To solve this, we can use the principle of conservation of energy, stating that the heat lost by the water will equal the heat gained by the ice cube as it warms up, melts, and then possibly continues to heat up as water. However, without knowing the specific heat capacities and the enthalpy of fusion, we cannot provide a numerical answer. Given the specific heats and the heat of fusion, one would set up an equation where the heat lost by the water equals the heat gained by the ice, and solve for the final temperature. The specific heat of water, the specific heat of ice, and the enthalpy of fusion of ice would be necessary information to perform actual calculations.

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        [tex]\Delta L=\frac{PL}{AE}[/tex]

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       [tex]A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2[/tex]

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       [tex]0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg[/tex]  

Mass of the climber = 69.38 kg

To find the diffusion rate in the second channel, we can use a formula that takes into account the cross-sectional area and length of the channels. Substituting the given values and simplifying the expression will give the diffusion rate in the second channel.

The diffusion rate is determined by several factors, including the concentration difference, the diffusion constant, temperature, and the size of the molecules. In this case, the diffusion rate in the first channel is given as 4.0 x 10^(-11) kg/s. To find the diffusion rate in the second channel, we can use the formula:

diffusion rate = (cross-sectional area of second channel / cross-sectional area of first channel) x (length of second channel / length of first channel) x diffusion rate of first channel

Substituting the given values:

diffusion rate = (0.30 cm^2 / 0.50 cm^2) x (0.10 cm / 0.25 cm) x (4.0 x 10^(-11) kg/s)

Simplifying the expression will give the diffusion rate in the second channel.

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The new diffusion rate in a channel with a cross-sectional area of 0.30 cm² and a length of 0.10 cm would be 6.0 x 10⁻¹¹ kg/s.

The diffusion rate for a solute in a solvent-filled channel is directly proportional to the cross-sectional area and inversely proportional to the length of the channel. Given that the initial diffusion rate is 4.0 x 10-11 kg/s in a channel with an area of 0.50 cm² and length 0.25 cm, to calculate the diffusion rate in a new channel with a different area and length, we use the equation:

Diffusion rate (new) = (Area (new) / Area (old)) x (Length (old) / Length (new)) x Diffusion rate (old)

Substituting the given values:

Diffusion rate (new) = (0.30 cm² / 0.50 cm²) x (0.25 cm / 0.10 cm) x 4.0 x 10⁻¹¹ kg/s

Diffusion rate (new) = (0.60) x (2.50) x 4.0 x 10⁻¹¹ kg/s

Diffusion rate (new) = 6.0 x 10⁻¹¹ kg/s

This calculation shows how the diffusion rate changes when altering the channel's dimensions.

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

[tex]\Delta L = L_f - L_i[/tex]

[tex]\Delta L = 21.5 - 11.9 = 9.6 cm[/tex]

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

[tex]mg = kx[/tex]

[tex]0.037 (9.81) = k(0.096)[/tex]

[tex]k = 3.78 N/m[/tex]

Now the period of oscillation of spring is given as

[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

Now plug in all values in it

[tex]T = 2\pi \sqrt{\frac{0.037}{3.78}}[/tex]

[tex]T = 0.62 s[/tex]

The period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.

We know that the formula of the spring force is written as,

[tex]F = k \times \delta[/tex]

Given to us,

weight on the spring, F = 37 g = (9.81 x 0.037) = 0.36297 N

Deflection in spring, δ = (21.5 - 11.9) = 9.6 cm = 0.096 m

Substitute the value,

[tex]0.36297 = k \times 0.096\\k= 3.78\rm\ N/m[/tex]

We know that the weight is been pulled down therefore, an external force is been applied to the spring to extend it further to 25.2 cm.

We know that the formula of the spring force is written as,

[tex]F = k \times \delta[/tex]

Given to us

Deflection in the spring, δ = (25.2 - 11.9) = 13.3 cm = 0.133 m

Spring constant, k = 3.78 N/m

Substitute the values,

[tex]F = 3.78 \times 0.133\\\\F = 0.5\\\\m \times g = 0.5\\\\m = 0.05125\rm\ kg[/tex]

We know the formula for the period of oscillation is given as,

[tex]T = 2\pi \sqrt{\dfrac{m}{k}}[/tex]

Substitute the values,

m = 0.05125 kg

k = 3.78 N/m

[tex]T = 2\pi \sqrt{\dfrac{0.05125}{3.78}}\\T = 0.7316\rm\ sec[/tex]

Hence, the period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.

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Answer:

Gravitational force = 89.18 x 10^-31 N

Electric force = 2.4 x 10^-17 N

Magnetic force = 7.68 x 10^-17 N

Explanation:

B = 80 micro tesla = 80 x 10^-6 T  north

E = 150 N/C downward

v = 6 x 10^6 m/s east

Gravitational force = m g = 9.1 x 10^-31 x 9.8 = 89.18 x 10^-31 N

Electric force = q E = 1.6 x 10^-19 x 150 = 2.4 x 10^-17 N

Magnetic force = q v B Sin 90 = 1.6 x 10^-19 x 6 x 10^6 x 80 x 10^-6

                          = 7.68 x 10^-17 N

Answer:

[tex]3.0\cdot 10^{-5} N/m^2[/tex]

Explanation:

The intensity of the radiation at the location of the satellite is given by:

[tex]I=\frac{P}{4\pi r^2}[/tex]

where

[tex]P=3.8\cdot 10^{26}W[/tex] is the power of the emitted radiation

[tex]4\pi r^2[/tex] is the area over which the radiation is emitted (the surface of a sphere), with

[tex]r=5.8\cdot 10^{10}m[/tex] being the radius of the orbit

Substituting,

[tex]I=\frac{3.8\cdot 10^{26}}{4\pi (5.8\cdot 10^{10})^2}=8989 W/m^2[/tex]

Now we can find the pressure of radiation, that for a totally absorbing surface (such as the satellite) is:

[tex]p=\frac{I}{c}[/tex]

where

[tex]c=3.0\cdot 10^8 m/s[/tex]

is the speed of light. Substituting,

[tex]p=\frac{8989}{3.0\cdot 10^8}=3.0\cdot 10^{-5} N/m^2[/tex]

Answer:

[tex]v_f = 20 m/s[/tex]

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

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

now for pure rolling condition we will have

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

also we have

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

now we will have

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

[tex]KE = mv^2[/tex]

now by work energy theorem we can say

[tex]W = KE_f - KE_i[/tex]

[tex]842 J = mv_f^2 - mv_i^2[/tex]

[tex]842 = 3(v_f^2) - 3\times 11^2[/tex]

now solve for final speed

[tex]v_f = 20 m/s[/tex]

Answer:

c) 359%

Explanation:

k = spring constant of the spring

x₁ = initial stretch of the spring = 7 cm = 0.07 m

x₂ = final stretch of the spring = 15 cm = 0.15 m

Percentage increase in energy is given as

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{((0.5)k{x_{2}}^{2} - {(0.5)k x_{1}}^{2})(100)}{(0.5)k{x_{1}}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({x_{2}}^{2} - {x_{1}}^{2})(100)}{{x_{1}}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({0.15}^{2} - {0.07}^{2})(100)}{{0.07}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}[/tex] = 359%

Explanation:

a) The equivalent resistance for resistors in series is the sum:

R = R₁ + R₂ + R₃

R = 10 Ω + 20 Ω + 30 Ω

R = 60 Ω

b) The resistors are in series, so they have the same current.

I = V/R

I = 12 V / 60 Ω

I = 0.2 A

c) The voltage drop can be found with Ohm's law:

V₁ = I R₁ = (0.2 A) (10 Ω) = 2 V

V₂ = I R₂ = (0.2 A) (20 Ω) = 4 V

V₃ = I R₃ = (0.2 A) (30 Ω) = 6 V

d) The power lost in each resistor is current times voltage drop:

P₁ = I V₁ = (0.2 A) (2 V) = 0.4 W

P₂ = I V₂ = (0.2 A) (4 V) = 0.8 W

P₃ = I V₃ = (0.2 A) (6 V) = 1.2 W

(a) The equivalent resistance of the circuit is 60 Ω.

(b) The current in each resistor is 0.2 A.

(c) The voltage across each resistor is 2 V, 4 V, and 6 V respectively.

(d) The power lost in each resistor is 0.4 W, 0.8 W and 1.2 W.

The equivalent resistance of the circuit is determined as follows;

Rt = R₁ + R₂ + R₃

Rt = 10 + 20 + 30

Rt = 60 Ω

The current in the resistors is calculated as follows;

V = IR

I = V/R

I = 12/60

I = 0.2 A

V1 =IR1 = 0.2 x 10 = 2 V

V2 = IR2  = 0.2 x 20 = 4 V

V3 = 0.2 x 30 = 6 V

P1 = I²R1 = (0.2)² x 10 = 0.4 W

P2 = I²R2 = (0.2)² x 20 = 0.8 W

P3 = I²R3 = (0.2)² x 30 = 1.2 W

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Answer:

Vector angular momentum about this axis of the sphere is:

L= 3.76[tex]\hat k[/tex] kg-m²/sec

Explanation:

The formula for the moment of inertia of a sphere is:

[tex]I=\frac{2}{5}\times MR^2[/tex]

Given:

Mass of the sphere = 13.5 kg

Radius of the sphere = 0.490 m

Thus, moment of inertia :

[tex]I=\frac{2}{5}\times 13.5\times (0.490)^2 kg\ m^2[/tex]

[tex]I=1.29654 kg\ m^2[/tex]

The expression for the angular momentum is:

L=I×ω

Given:

Angular speed(ω) = 2.9 rad/s

I, above calculated = 1.29654 kgm⁻²

Thus, angular momentum is:

L= 1.29654×2.9 kg-m²/sec

L= 3.76 kg-m²/sec

Given, the sphere is turning counterclockwise about the vertical axis. Thus, the direction of the angular momentum will be on the upper side of the plane. ( [tex]+\hat k[/tex] ).

Thus, angular momentum with direction is:

L= 3.76[tex]\hat k[/tex] kg-m²/sec

Answer:

amplitude, a = 3.076 mm

frequency, f = 10.35 Hz

Explanation:

Vmax = 200 mm /s = 0.2 m/s

Amax = 13 m/s

Let f be the frequency and a be the amplitude.

Use the formula of maximum velocity.

Vmax = ω a

0.2 = ω a    ..... (1)

Use the formula of maximum acceleration.

Amax = ω^2 x a

13 = ω^2 a    ..... (2)

Divide equation (2) by (1)

ω = 13 / 0.2 = 65 rad/s

Put in equation (1)

0.2 = 65 x a

a = 3.076 x 10^-3 m

a = 3.076 mm

Let f be the frequency

ω = 2 π f

f = 65 / (2 x 3.14) = 10.35 Hz  

Answer:

0.3 N

Explanation:

mass of cork = 0.01 kg, density of cork = 250 kg/m^3

density of water = 1000 kg/m^3, g = 10 m/s^2

Tension in the rope = Buoyant force acting on the cork - Weight of the cork

Buoyant force = volume of cork x density of water x g

                        = mass x density of water x g / density of cork

                       = 0.01 x 1000 x 10 / 250 = 0.4 N

Weight of cork = mass of cork x g = 0.01 x 10 = 0.1 N

Thus, the tension in the rope = 0.4 - 0.1 = 0.3 N

Answer:

Tension = 0.3 N

Explanation:

As we know that the cork is inside water

so the buoyancy force on the cork is counter balanced by tension force in string and weight of the block

So the force equation is given as

[tex]F_b = T + mg[/tex]

now we will have

[tex]Volume = \frac{mass}{density}[/tex]

[tex]V = \frac{0.01}{250} = 4 \times 10^{-5} m^3[/tex]

now buoyancy force on the block is given by

[tex]F_b = \rho V g[/tex]

[tex]F_b = 1000(4 \times 10^{-5})(10)[/tex]

[tex]F_b = 0.4 N[/tex]

now by force balance equation

[tex]0.4 = T + 0.01(10)[/tex]

[tex]T = 0.4 - 0.1 = 0.3 N[/tex]

Answer:

(a) 9.65 mH

(b) doubled

Explanation:

A = 6 cm^2 = 6 x 10^-4 m^2, l = 20 cm = 0.2 m, i = 2.8 A, N = 1600

(a) The formula for the self inductance of a solenoid is given by

L = μ0 N^2 A / l

L = 4 x 3.14 x 10^-7 x 1600 x 1600 x 6 x 10^-4 / 0.2

L = 9.65 x 10^-3 H

L = 9.65 mH

(b) As we observe that the inductance is proportional to the area so if the area is doubled, then inductance is also doubled.

Answer:

Explanation: r₂ = 15 x 10⁻³ m ; r = 12 x 10⁻³m ; r₁ = 10 x 10⁻³ m

     E = Q /4π∈r²x ( r³ - r₁ ³) / ( r₂³ - r₁³ )

         = 15 x 10⁻⁶ x 8.99 x 10⁹x( 12³ - 10³ ) / ( 15³- 10³ ) X 12³

          = 23.92 N/C

The electric field is the force acting on a charge located in a region in space. The electric field at a distance r from the center of the shell is 2.8x10⁶N/C

The electric field (E) can be defined as a region in the space that interacts with electric charges or charged bodies through electric forces.

It is a vectorial field in which an electric charge (q) suffers the effect of the electric force (F). In other words, it represents the force that is acting on a charge located in a certain region.

E = F/q

It is expressed in newton/coulomb (N/C)

In the exposed example, we have the following data,

We need to get the value of the electric field at the given distance r.

E = Q/4πεr² = kQ/r²

Since the r value is located between r1 and r2, we need to get the electric field at that intermedia distance, r. This is a new ratio.

The total charge Q is distributed in the whole area of the shell thickness or density, delimited by r1 and r2.

But to get the electric field at the intermedia distance, r, we only need to get the charge q lower than Q and distributed between the r and r1, which represents a volume v lower than the total volume V of the shell.

So to get the q value, we need to consider the charge volumetric density, σ.

σ = Q/V

Where

But to get Q/V, we need to get the total shell volume value, V.

V = 4/3 π r³

V = 4/3 π (r2³ - r1³)

V = 4/3 π (0.15³ - 0.1³)

V = 4/3 π (0.00237)

V = 0.00993m³

Now that we have the total volume, and we know the value of the total charge, we need to get the density of the shell.

σ = Q/V

σ = 15 μC / 0.00993m³

σ = 15x10⁻⁶ C / 0.00993m³

σ = 0.00151 C/m³

Now, we will calculate the volume v between r and r1, which is what we are interested in.

v = 4/3 π r³

v = 4/3 π (r³ - r1³)

v = 4/3 π (0.12³ - 0.1³)

v = 4/3 π (0.0007)

v = 0.00305m³

Finally, using this new v value and the σ value, we can get the charge q distributed  throughout the volume v.

q = vσ

q = 0.00305m³ x 0.00151 C/m³

q = 4.6x10⁻⁶ C

Now we can get the electric field at a distance r from the center of the shell

E = kq/r²

E = ((8.99 × 10⁹ N ∙ m²/C²) x (4.6x10⁻⁶ C) ) / 0.12²m

E = 2.8x10⁶N/C

Hence, the electric field at a distance r is 2.8x10⁶N/C

You can learn more about electric field at

https://brainly.com/question/8971780

Answer:

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

[tex]a_{new} = 8a.[/tex]

Explanation:

given data:

radius of orbit = R

Speed pf planet = v

new radius = 0.5R

new speed = 2v

we know that acce;ration is given as

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

[tex]a_{new} =\frac{(2v)^{2}}{0.5R},[/tex]

           [tex]= \frac{4v^{2}}{0.5R}[/tex]

          [tex]= \frac{8 v^{2}}{R}[/tex]

[tex]a_{new} = 8a.[/tex]

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

The magnitude of the probe’s acceleration in the new orbit is 8 times the magnitude of the probe’s acceleration in the old orbit.

The acceleration acted on the body moving in a closed circular path towards the center of the curve is known as centripetal acceleration. Due to centripetal acceleration, the body is able to move in a closed circular path.

[tex]a_C=\frac{V^{2} }{R}[/tex]

For old orbit

the radius of orbit = R

Speed pf planet = v

[tex]a_C=\frac{V^{2} }{R}[/tex]

For new orbits

new radius = 0.5R

new speed = 2v

[tex]a_C=\frac{(2V)^{2} }{0.5R}[/tex]³

[tex]a_C=\frac{8V^{2} }{R}[/tex]

Hence,

[tex]a_C_{new}=8a_C__{old}[/tex]

The magnitude of the probe’s acceleration in the new orbit is 8 times the magnitude of the probe’s acceleration in the old orbit.

To learn more about centripetal acceleration refers to the link

https://brainly.com/question/17689540