The electric potential of a charge distribution is given by the equation V(x) = 3x2y2 + yz3 - 2z3x, where x, y, z are measured in meters and V is measured in volts. Calculate the magnitude of the electric field vector at the position (x,y,z) = (1.0, 1.0, 1.0)

Answer:

ΔL = 1.653 km

Explanation:

The linear expansion of any object due to change in temperature is given by the following formula:

ΔL = αLΔT

where,

ΔL = Change in length or expansion of steel pipe line = ?

α = coefficient of linear expansion of steel = 12 x 10⁻⁶ /°C

L = Original Length of the steel pipe = 1300 km

ΔT = Change in temperature = 35°C - (- 71°C) = 35°C + 71°C = 106°C

Therefore,

ΔL = (12 x 10⁻⁶ /°C)(1300 km)(106°C)

ΔL = 1.653 km

Answer:

no two electrons in the same atom can have the same set of four quantum numbers

Explanation:

Pauli 's Theory of Exclusion specifies that for all four of its quantum numbers, neither two electrons in the same atom can have similar value.

In a different way, we can say that no more than two electrons can take up the identical orbital, and two electrons must have adversely spin in the identical orbital

Therefore the second option is correct

Answer:

Both attempt to explain human behavior

Explanation:

Psychology is generally regarded as the science of human behavior. Behaviourism is the psychological theory which holds that behaviour can be fully understood in terms of conditioning, without actually considering thoughts or feelings. The theory holds that psychological disorders can be aptly handled by simply altering the behavioural patterns of the individual. It involves the study of stimulus and responses.

Cognitive psychology attempts to decipher what is going on in people's minds. That is, it looks at the mind as a processor of information. Hence we can define cognitive psychology as the study of the internal mental processes. This according to behaviorists, cannot be studied in measurable terms as in behaviourism (stimulus response approach) even though mental processes are known to influence human behavior significantly.

Hence, both behaviourism and cognitive psychology attempt to study human behavior from different perspectives.

Answer:

0.218

Explanation:

Given that

Total vibrations completed by the wave is 43 vibrations

Time taken to complete the vibrations is 33 seconds

Length of the wave is 424 cm = 4.24 m

to solve this problem, we first find the frequency.

Frequency, F = 43 / 33 hz

Frequency, F = 1.3 hz

Also, we find the wave velocity. Which is gotten using the relation,

Wave velocity = 4.24 / 15

Wave velocity = 0.283 m/s

Now, to get our answer, we use the formula.

Frequency * Wavelength = Wave Velocity

Wavelength = Wave Velocity / Frequency

Wavelength = 0.283 / 1.3

Wavelength = 0.218

Answer:

Yes we can induce current in the coil by moving the magnet in and out of the coil steadily.

Explanation:

A current can be induced there using the magnetic field and the coil of wire. Moving the bar magnet around the coil can induce a current and this is called electromagnetic induction.

The generation of an electromotive force  across an electrical conductor in a fluctuating magnetic field is known as electromagnetic or magnetic induction.

Induction was first observed in 1831 by Michael Faraday, and James Clerk Maxwell mathematically named it Faraday's law of induction. The induced field's direction is described by Lenz's law.

Electrical equipment like electric motors and generators as well as parts like inductors and transformers have all found uses for electromagnetic induction.

Here, moving the bar magnet around the coil generates the electronic movement followed by a generation of electric current.

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Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

[tex]\Delta P \propto \rho \cdot \Delta h[/tex]

[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]

Where:

[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.

[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.

[tex]\Delta h[/tex] - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]

[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]

Where:

[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.

[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.

If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:

[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]

[tex]\Delta h_{air} = 158.140\,m[/tex]

The height of the building is 158.140 meters.

158.13m

Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e

P = ρhg

Let;

Pressure measured at the roof top =  ([tex]P_{R}[/tex])

Pressure measured at the ground level =  ([tex]P_{G}[/tex])

Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P               ---------------(i)

(a) P = ρhg             -----------(***)

But;

ρ = density of air = 1.29kg/m³  

h = height of column of air = height of building

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (***)

P = 1.29 x h x 10

P = 12.9h Pa

(b) [tex]P_{G}[/tex] =  ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (*)

[tex]P_{G}[/tex] =  13600 x 0.76 x 10

[tex]P_{G}[/tex] = 103360 Pa

(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g       --------------(**)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (**)

[tex]P_{R}[/tex]  =  13600 x 0.745 x 10

[tex]P_{R}[/tex]  = 101320 Pa

(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P  

103360 = 101320 + 12.9h

Solve for h;

12.9h = 103360 - 101320

12.9h = 2040

h = [tex]\frac{2040}{12.9}[/tex]

h = 158.13m

Therefore, the height of the building is 158.13m

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

[tex]\triangle E = 12.79 J[/tex]

Explanation:

Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]

Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]

Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]

Difference in maximum stored energy between the sprinters and the non-athlethes:

[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]

Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

Answer:

( About ) 6.8nC

Explanation:

We are given the equation |Q2| = mgd^2 / kQ1. Let us substitute known values into this equation, but first list the given,

Charge Q2 = +45nC = (45 × 10⁻⁹) C

mass of charge Q2 = 4.5 μg, force of gravity = 4.5 μg × 9.8 m/s² = ( 4.41 × 10^-5 ) N,

Distance between charges = 25 cm = 0.25 m,

k = Coulomb's constant = 9 × 10^9

_______________________________________________________

And of course, we have to solve for the magnitude of Q2, represented by the charge magnitude of the charge on Q2 -

(4.41 × 10^-5) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] / 0.25²

_______________________________________________________

Solution = ( About ) 6.8nC

Answer:

F₂ = 925.92 N

Explanation:

In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,

σ₁ = σ₂

F₁/A₁ = F₂/A₂

F₂ = F₁ A₂/A₁

where,

F₂ = Initial force that must be applied to narrow arm = ?

F₁ = Load on Wider Arm to be raised = 12000 N

A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²

A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²

Therefore,

F₂ = (12000 N)(25π cm²)/(324π cm²)

F₂ = 925.92 N

Answer:

A charged atom or particle is called an ion :)

Answer:

20 J

Explanation:

From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.

m'u'+mu = V(m+m')........... Equation 1

Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.

make V the subject of the equation above

V = (m'u'+mu)/(m+m')............. Equation 2

Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).

Substitute into equation 2

V = [(2×5)+(8×0)]/(2+8)

V = 10/10

V = 1 m/s.

Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision

Total kinetic energy before collision = 1/2(2)(5²) = 25 J

Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J

Lost in kinetic energy = 25-5 = 20 J

The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.

Since there is a lost in kinetic energy, the collision is inelastic collision.  

m'u'+mu = V(m+m')

[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]  

Where

m' = mass of the moving object = 2 kg

m = mass of the stick = 8 kg,

u' = initial velocity of the moving object = 5 m/s

V = common velocity after collision= ?    

u = 0 m/s (at rest).

put the values in the formula,  

[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]

  kinetic energy before collision

[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]  

kinetic energy after collision

[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]  

Lost in kinetic energy = 25-5 = 20 J

Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.

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Answer:

a) vo = 14m/s

b) v = 14m/s

c) t = 2.85s

d) t = 0.829s

e) v =  22.12 m/s

Explanation:

a) To find the initial velocity of the ball yo use the following formula:

[tex]h_{max}=\frac{v_o^2}{2g}[/tex]         (1)

hmax:  maximum height reached by the ball but measured from the point at which the ball is thrown = 25.0m - 15.0m = 10.0m

vo: initial velocity of the ball = ?

g: gravitational acceleration = 9.8m/s^2

You solve the equation (1) for vo and replace the values of the other parameters:

[tex]v_o=\sqrt{2gh_{max}}}=\sqrt{2(9.8m/s^2)(10.0m)}=14\frac{m}{s}[/tex]

The initial velocity of the ball is 14m/s

b) To find the velocity of the ball when it is at the same position as the initial point where it was thrown, you can use the following formula:

[tex]v^2=2gh_{max}\\\\v=\sqrt{2gh_{max}}[/tex]        

as you can notice, v = vo = 14m/s

The velocity of the ball is 14 m/s

c) The flight time of the ball is given by twice the time the ball takes to reach the maximum height. You use the following formula:

[tex]t=2\frac{v_o}{g}=2\frac{14m/s}{9.8m/s^2}=2.85s[/tex]             (3)

The time is 2.85s

d) To find the time the ball takes to arrive to the ground after the ball passes the same height at which is was thrown, you can use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]          (4)

y: 0 m (ball just after it impact the ground)

yo: initial position = 15.0 m

vo: in)itial velocity of the ball = 14m/s    

t: time

You replace the values of the parameters in the equation (4) and obtain a quadratic formula:

[tex]0=15.0-14t-\frac{1}{2}(9.8)t^2\\\\[/tex]

You use the quadratic formula to find the roots t:

[tex]t_{1,2}=\frac{-(-14)\pm\sqrt{(-14)^2-4(4.9)(15)}}{2(-4.9)}\\\\t_{1,2}=\frac{14\pm22.13}{-9.8}\\\\t_1=0.829s\\\\t_2=-2.19s[/tex]

you choose the positive values because is has physical meaning

The time the ball takes to arrive to the ground is 0.829s

e) The final velocity is:

[tex]v=v_o+gt[/tex]

[tex]v=14m/s+(9.8m/s^2)(0.829s)=22.12\frac{m}{s}[/tex]

The final velocity is 22.14 m/s

Answer:

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Explanation:

An electron has a mass of [tex]9.1 \times 10^{-31}\,kg[/tex] and a charge of [tex]-1.6 \times 10^{-19}\,C[/tex]. Based on the Principle of Charge Conservation, [tex]-2.10\times 10^{-6}\,C[/tex] in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:

[tex]n_{R} = \frac{-2.10\times 10^{-6}\,C}{-1.6 \times 10^{-19}\,C}[/tex]

[tex]n_{R} = 1.3125 \times 10^{13}\,electrons[/tex]

The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:

[tex]n_A = \frac{m_{ball}}{M_{Cu}}\cdot N_{A}[/tex]

Where:

[tex]m_{ball}[/tex] - Mass of the ball, measured in kilograms.

[tex]M_{Cu}[/tex] - Atomic mass of cooper, measured in grams per mole.

[tex]N_{A}[/tex] - Avogradro's Number, measured in atoms per mole.

If [tex]m_{ball} = 56\,g[/tex], [tex]M_{Cu} = 63.5\,\frac{g}{mol}[/tex] and [tex]N_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}[/tex], the number of atoms is:

[tex]n_{A} = \left(\frac{56\,g}{63.5\,\frac{g}{mol} } \right)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)[/tex]

[tex]n_{A} = 5.3107\times 10^{23}\,atoms[/tex]

As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:

[tex]n_{E} = \left(29\,\frac{electrons}{atom} \right)\cdot (5.3107\times 10^{23}\,atoms)[/tex]

[tex]n_{E} = 1.5401\times 10^{23}\,electrons[/tex]

The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:

[tex]x = \frac{n_{R}}{n_{E}}[/tex]

[tex]x = \frac{1.3125\times 10^{13}\,electrons}{1.5401\times 10^{23}\,electrons}[/tex]

[tex]x = 8.5222 \times 10^{-11}[/tex]

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Answer:

Please, read the anser below

Explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

[tex]F_2-F_1-F_f=Ma[/tex]          (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

[tex]a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}[/tex]

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero

Answer:

a

  [tex]PE = 2.3 *10^{-16} \ J[/tex]

b

 [tex]T = 1.1 *10^{7} \ K[/tex]

Explanation:

From the question we are told that

      The distance of separation is  [tex]d = 1.00 *10^{-12} \ m[/tex]

Generally the electric potential energy can be mathematically represented as

            [tex]PE = \frac{k * q_1 q_2 }{d}[/tex]

Given that in a nuclei the only charged particle is the proton who charge is

     [tex]p = 1.60 *10^{-19} \ C[/tex]

Hence

     [tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]

And k is the coulomb constant with values   [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]

      So we have that

       [tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]

      [tex]PE = 2.3 *10^{-16} \ J[/tex]

The relationship between the electrical potential energy and the temperature is mathematically represented as

         [tex]PE = \frac{3}{2} kT[/tex]

Here  k is  the Boltzmann's constant with value  [tex]k = 1.38*10^{-23} JK^{-1}[/tex]

   making T the subject

       [tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]

substituting values

      [tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]

     [tex]T = 1.1 *10^{7} \ K[/tex]

Answer:

a) [tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex], b) Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex], c) The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].

Explanation:

a) The total energy of the object is equal to the sum of potential and kinetic energies. That is:

[tex]E = K + U[/tex]

Where:

[tex]K[/tex] - Kinetic energy, dimensionless.

[tex]U[/tex] - Potential energy, dimensionless.

After replacing each term, the total energy of the object at any point in its motion is:

[tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex]

b) The amplitude of the motion occurs when total energy is equal to potential energy, that is, when objects reaches maximum or minimum position with respect to position of equilibrium. That is:

[tex]E = U[/tex]

[tex]E = \frac{1}{2} \cdot k \cdot A^{2}[/tex]

Amplitude is finally cleared:

[tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex]

Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex].

c) The maximum speed of the motion when total energy is equal to kinetic energy. That is to say:

[tex]E = K[/tex]

[tex]E = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]

Maximum speed is now cleared:

[tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex]

The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].

Answer:

P = 5.97 × 10^(5) Pa

Explanation:

We are given;

Mass of balloon;m_b = 4.3 kg

Radius;r = 1.54 m

Temperature;T = 289 K

Density;ρ = 1.19 kg/m³

We know that, density = mass/volume

So, mass = Volume x Density

We also know that Force = mg

Thus;

F = mg = Vρg

Where m = mass of balloon(m_b) + mass of helium (m_he)

So,

(m_b + m_he)g = Vρg

g will cancel out to give;

(m_b + m_he) = Vρ - - - eq1

Since a sphere shaped balloon, Volume(V) = (4/3)πr³

V = (4/3)π(1.54)³

V = 15.3 m³

Plugging relevant values into equation 1,we have;

(3 + m_he) = 15.3 × 1.19

m_he = 18.207 - 3

m_he = 15.207 kg = 15207 g

Molecular weight of helium gas is 4 g/mol

Thus, Number of moles of helium gas is ; no. of moles = 15207/4 ≈ 3802 moles

From ideal gas equation, we know that;

P = nRT/V

Where,

P is absolute pressure

n is number of moles

R is the gas constant and has a value lf 8.314 J/mol.k

T is temperature

V is volume

Plugging in the relevant values, we have;

P = (3802 × 8.314 × 289)/15.3

P = 597074.53 Pa

P = 5.97 × 10^(5) Pa

Answer:

The answer is electromagnetic

Answer:

electromagnetic

Explanation:

edge 2021

Answer: According to Newtons second Law of motion ;

F = ma (Force  equals  mass multiplied by acceleration.)

The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force

Explanation:

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

P.E = 0.068 J = 68 mJ

Answer:

[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]

Explanation:

Given that

Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]

slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]

slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]

We know that for double slit the maxima condition is that

[tex]\operatorname{dsin} \theta=m \lambda[/tex]

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

[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]

For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]

[tex]\tan \theta=\frac{y_{m}}{D}[/tex]

[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]

Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]

Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]

Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, [tex]V_s[/tex] = 450 V

potential at radial distance, [tex]V_r[/tex] = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]

[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]

Therefore, the radius of the sphere is 4.05 m

Answer:

Force

Explanation:

Force is simply known as pull or push of an object

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

Answer:

a) v₁ = 3.92 m / s , b)     ΔP =  = 9.0 10⁴ Pa, c)  t = 0.0297 s  

Explanation:

This is a fluid mechanics exercise

a) let's use the continuity equation

let's use index 1 for the hose and index 2 for the nozzle

        A₁ v₁ = A₂v₂

in area of ​​a circle is

       A = π r² = π d² / 4

we substitute in the continuity equation

        π d₁² / 4 v₁ = π d₂² / 4 v₂

        d₁² v₁ = d₂² v₂

the speed of the water in the hose is v1

       v₁ = v₂ d₂² / d₁²

       v₁ = 14 (1.8 / 3.4)²

        v₁ = 3.92 m / s

b) they ask us for the pressure difference, for this we use Bernoulli's equation

       P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2

as the hose is horizontal y₁ = y₂

       P₁ - P₂ = ½ ρ (v₂² - v₁²)

      ΔP = ½ 1000 (14² - 3.92²)

       ΔP = 90316.8 Pa = 9.0 10⁴ Pa

c) how long does a tub take to flat

the continuity equation is equal to the system flow

        Q = A₁v₁

        Q = V t

where V is the volume, let's equalize the equations

         V t = A₁ v₁

         t = A₁ v₁ / V

A₁ = π d₁² / 4

let's reduce it to SI units

         V = 120 l (1 m³ / 1000 l) = 0.120 m³

          d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m

let's substitute and calculate

         t = π d₁²/4   v1 / V

         t = π (3.4 10⁻²)²/4 3.92 / 0.120

         t = 0.0297 s

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

[tex]P = \frac{F}{A}[/tex]

Hydraulic Lift - After change

[tex]P + \Delta P = \frac{F + \Delta F}{A}[/tex]

Where:

[tex]P[/tex] - Hydrostatic pressure, measured in pascals.

[tex]\Delta P[/tex] - Change in hydrostatic pressure, measured in pascals.

[tex]A[/tex] - Cross sectional area of the hydraulic lift, measured in square meters.

[tex]F[/tex] - Hydrostatic force, measured in newtons.

[tex]\Delta F[/tex] - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

[tex]\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}[/tex]

[tex]\Delta P = \frac{\Delta F}{A}[/tex]

[tex]\Delta F = A\cdot \Delta P[/tex]

Given that [tex]\Delta P = 100\,Pa[/tex] and [tex]A = 25\,m^{2}[/tex], the additional weight is:

[tex]\Delta F = (25\,m^{2})\cdot (100\,Pa)[/tex]

[tex]\Delta F = 2500\,N[/tex]

The additional mass needed for the additional weight is:

[tex]\Delta m = \frac{\Delta F}{g}[/tex]

Where:

[tex]\Delta F[/tex] - Additional weight, measured in newtons.

[tex]\Delta m[/tex] - Additional mass, measured in kilograms.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

If [tex]\Delta F = 2500\,N[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:

[tex]\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\Delta m = 254.92\,kg[/tex]

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer:

The magnetic field through the coil at first increases steadily up to its maximum value, and then decreases gradually to its minimum value.

Explanation:

At first, the magnet fall towards the coils;  inducing a gradually increasing magnetic field through the coil as it falls into the coil. At the instance when half the magnet coincides with the coil, the magnetic field magnitude on the coil is at its maximum value. When the magnet falls pass the coil towards the floor, the magnetic field then starts to decrease gradually from a strong magnitude to a weak magnitude.

This action creates a changing magnetic flux around the coil. The result is that an induced current is induced in the coil, and the induced current in the coil will flow in such a way as to oppose the action of the falling magnet. This is based on lenz law that states that the induced current acts in such a way as to oppose the motion or the action that produces it.