The emf of a battery is equal to its terminal potential difference: A) under all conditions B) only when the battery is being charged C) only when a large cwrrent is in the battery D) only when there is no current in the battery E) under no conditions

Answer:

So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical

Explanation:

When two polarizers are arranged at 45 degree and 90 degree in series

so here the intensity of light coming out of the polarizers is given as

[tex]I_1 = I_ocos^45[/tex]

[tex]I_1 = \frac{I_0}{2}[/tex]

now for second polarizer we have

[tex]I_2 = I_1 cos^2(90 - 45)[/tex]

[tex]I_2 = (\frac{I_0}{2})\frac{1}{2}[/tex]

[tex]I_2 = \frac{I_0}{4}[/tex]

now in other case when two polarizers are inclined 90 degree to the vertical

[tex]I = I_ocos^290 = 0[/tex]

so final intensity in second case will be ZERO

So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical

Answer:

F = 18195.59 N or F = 18196 rounded up

Explanation:

force = GMm/d^2

G = 6.67x10^-11

M = 5.98x10^24 kg

m = 2170kg

d = 6370000 + 527000 = 6897000m

putting all values   (6.67x10^-11)(5.98x10^24)(2170)/(6897000^2) = 18195.59....

The magnitude of the force with which the space station attracts Earth is about 1.82 × 10⁴ Newton

[tex]\texttt{ }[/tex]

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of space station = m = 2170 kg

radius of the orbit = R = 5.27 × 10⁵ + 6.37 × 10⁶ = 6.897 × 10⁶ m/s

mass of Earth = M = 5.98 × 10²⁴ kg

Asked:

Gravitational Force = F = ?

Solution:

[tex]F = G \frac{M.m}{R^2}[/tex]

[tex]F = 6.67 \times 10^{-11} \times \frac{5.98 \times 10^{24} \times 2170}{(6.897 \times 10^6)^2}[/tex]

[tex]F \approx 1.82 \times 10^4 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer and Explanation:

The two necessary design conditions before an ion chamber can be considered a Bragg-Grey chamber are:

Such a chamber can be used for the purpose of calculation and measurement of the absorbed ionizing radiation dose inside the chamber. Basically, ionization chambers are used as detectors for X-ray absorption spectroscopy.

Answer:

The absolute potential 4.0 m away from the same point charge is -50 V.

(A) is correct option.

Explanation:

Given that,

Distance = 2.0 m

Potential = -100 V

Absolute potential = 4.0 m

We need to calculate the charge

Using formula of potential

[tex]V=\dfrac{kq}{r}[/tex]

Where, V = potential

q = charge

r = distance

Put the value into the formula

[tex]-100=\dfrac{9\times10^{9}\times q}{2.0}[/tex]

[tex]q=\dfrac{200}{9\times10^{9}}[/tex]

[tex]q=-22.2\times10^{-9}\ C[/tex]

We need to calculate the potential

Using formula of potential

[tex]V=\dfrac{9\times10^{9}\times(-22.2\times10^{-9})}{4.0}[/tex]

[tex]V=-49.95\ V[/tex]

[tex]V=-50\ V[/tex]

Hence, The absolute potential 4.0 m away from the same point charge is -50 V.

Answer:

6.6 x 10^5 Nm^2/C

Explanation:

E = 125000 N/C

Area, A = length x width = 2.5 x 5 = 12.5 m^2

θ = 65 degree

electric flux, φ = E A Cosθ

φ = 125000 x 12.5 x Cos 65

φ = 6.6 x 10^5 Nm^2/C

Here, we are required to evaluate the electric flux through the rectangle.

The electric flux through the rectangle is;

φ = 6.6 × 10^5 Nm²/C

The electric flux through a body is given as;

electric flux, φ = E A Cosθ

where;.

The Area of the rectangle, A = 2.5 × 5

Therefore, A = 12.5m².

Therefore, the electric flux, φ = E A Cosθ

φ = 125000 × 12.5 × Cos65

φ = 660,341 Nm²/C.

φ = 6.6 × 10^5 Nm²/C

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Answer: a) [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n[/tex]

b) [tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}[/tex]

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n[/tex]

Now,  as the mass on both reactant and product side must be equal:

[tex]235+1=87+146+x[/tex]

[tex]x=3[/tex]

Thus three neutrons are produced and nuclear equation will be: [tex]^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n[/tex]

b) For the another fission reaction:

[tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}[/tex]

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

[tex]^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}[/tex]

Final answer:

The fission of uranium-235 can yield multiple products including bromine-87, lanthanum-146, and additional neutrons. Part a explains the specific fission process forming these products, while part b outlines a scenario producing a nucleus with 56 protons and another with 94 nucleons, alongside more neutrons.

Explanation:

The question relates to the process of nuclear fission, specifically the fission of uranium-235 (U-235) when it captures a neutron. Nuclear fission involves a large nucleus splitting into smaller nuclei along with the release of energy and additional neutrons. These neutrons can initiate further fission reactions, creating a chain reaction.

For part a (forming bromine-87 and lanthanum-146):

U-235 + n → Br-87 + La-146 + 3n

This equation shows that upon absorbing a neutron, U-235 splits into bromine-87, lanthanum-146, and releases 3 more neutrons.

For part b (forming a nucleus with 56 protons and another with a total of 94 nucleons):

A nucleus with 56 protons corresponds to iron (Fe), due to the atomic number. A second nucleus having a total of 94 nucleons with an unspecified proton-to-neutron ratio could refer to an element like plutonium-94, considering the total number of protons and neutrons. However, the provided information primarily focuses on the fission products rather than specifics on this second nucleus. Additionally, 2 more neutrons are produced. A precise equation for this scenario cannot be formulated without more specific information on the second fragment.

Answer:

[tex]N = 3[/tex]

so these are

either 3 to 2 or 2 to 1 or 3 to 1

Explanation:

As we know that total number photons emitted from nth state to ground state is given by

[tex]N = ^nC_2[/tex]

[tex]N = \frac{n(n-1)}{2}[/tex]

here it shows that if electron has transition from nth excited state to ground state then total number of photons is given by above equation

here we know that higher state is n = 3

so total number of photons is given as

[tex]N = \frac{3(3-1)}{2}[/tex]

[tex]N = 3[/tex]

so these are

either 3 to 2 or 2 to 1 or 3 to 1

Answer:

42 m

Explanation:

[tex]v_{o}[/tex] = initial velocity of the car = 24 m/s

[tex]v_{f}[/tex] = final velocity of the car = 0 m/s

μ = coefficient of kinetic friction = 0.7

g = acceleration due to gravity = 9.8 m/s²

a = acceleration due to kinetic frictional force  = - μg = - (0.7)(9.8) = - 6.86 m/s²

d = distance through which the car skids

Using the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d[/tex]

Inserting the values

[tex]0^{2} = 24^{2} + 2 (- 6.86) d[/tex]

d = 42 m

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u =  3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0² = 2.98² + 2 x -9.81 x s

    s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Answer : The atomic weight of the element is, 121.75 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

As we are given that,

Mass of isotope 1 = 120.9038 amu

Percentage abundance of isotope 1 = 57.25 %

Fractional abundance of isotope 1 = 0.5725

Mass of isotope 2 = 122.8831 amu

Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %

Fractional abundance of isotope 2 = 0.4275

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

[tex]\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)][/tex]

[tex]\text{Average atomic mass of element}=121.75amu[/tex]

Therefore, the atomic weight of the element is, 121.75 amu

The atomic weight of an element with two isotopes of 120.9038 amu at 57.25% abundance and 122.8831 amu is calculated as 121.75 amu after rounding to two decimal places.

The atomic weight of an element with two isotopes can be calculated using the weighted average formula. This takes into account the atomic masses of the isotopes and their relative abundances.

Since one isotope has an atomic weight of 120.9038 amu with 57.25% abundance and we know the atomic weight of the other isotope is 122.8831 amu, we can calculate the average atomic mass as follows:

The percent abundance of the second isotope must be 100% - 57.25% = 42.75%, since the two abundances must add up to 100%.

Atomic weight = (isotope1_mass * isotope1_abundance) + (isotope2_mass * isotope2_abundance)

Substituting our values in gives us:
Atomic weight = (120.9038 amu * 0.5725) + (122.8831 amu * 0.4275)

Atomic weight = (69.1866935 amu) + (52.53009825 amu)

Atomic weight = 121.71679175 amu

After rounding to two decimal places, the average atomic weight of the element is 121.72 amu which is closest to option 4, 121.75 amu.

Hey there!

a) The electric field is in direction of decreasing value of potential so the electric field here will points towards 0v from 9 v. But the charge on particle is negative so the force will be towards the 9v point. and so the charge will move towards the 9 V point.

b) Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location while.the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. Electric potential is always a relative quantity because no one can find out absolute potential at a point, so in reality only potential difference exists. We can assume potential at certain point (say at infinity it is zero) only then we are able to define potential at every point.

c) The knowlendge of electric potential helps us in finding the work done on the particle which is used in work energy theorem to find out the chanrge in kinetic energy and other stuff.

Hope this helps!

Answer:

3888.23 N

Explanation:

m = mass of the person = 78.03 kg

g = standard acceleration due to gravity = 8.23 m/s²

a = acceleration of the aircraft = 5.05 g = 5.05 x 8.23 = 41.6 m/s²

F = upward force the person experience

Force equation for the motion of person is given as

F - mg = ma

Inserting the values

F - (78.03)(8.23) = (78.03)(41.6)

F = 3888.23 N

Answer:

0.25 T

Explanation:

F = Force required to keep the bar moving = 0.080 N

B = magnitude of magnetic field = ?

L = length of the bar = 50 cm = 0.50 m

v = speed of the bar = 0.50 m/s

R = resistance of the bar =0.10 Ω

Force is given as

[tex]F = \frac{B^{2}L^{2}v}{R}[/tex]

[tex]0.08 = \frac{B^{2}(0.50)^{2}(0.50)}{0.10}[/tex]

B = 0.25 T

To find the magnitude of the magnetic field, use the equation F = qvBsinθ and Ohm's Law to solve for q. Then substitute q back into the equation to find the magnetic field.

To find the magnitude of the magnetic field, we can use the equation F = qvBsinθ. In this case, the force (F) is given as 0.080 N, the charge (q) is not given but is not needed to find the magnetic field, the velocity (v) is given as 0.50 m/s, and sinθ is 1 because the angle between the velocity and the magnetic field is 90°. Rearranging the equation to solve for B, we get B = F / (qv sinθ). Substituting the given values, we have B = 0.080 N / (q * 0.50 m/s * 1). The only unknown variable left is the charge (q).

To calculate the charge, we can use the given resistance and Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance. Rearranging the equation to solve for I, we get I = V / R. The voltage across the resistor can be calculated using the formula V = Fd, where d is the distance between the rails, which is given as 50 cm (or 0.5 m). Substituting the given values, we have V = 0.080 N * 0.5 m. Now, we can substitute the calculated current (I) into the equation B = 0.080 N / (q * 0.50 m/s * 1) and solve for q. Finally, we can substitute the calculated charge (q) back into the equation B = 0.080 N / (q * 0.50 m/s * 1) to find the magnitude of the magnetic field.

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Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

[tex]\frac{dA}{dt}[/tex] = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B [tex]\frac{dA}{dt}[/tex]

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

Answer:

2710.66 V

Explanation:

Eo = 90 V, R = 21.1 ohm, L = 1.05 H, C = 2.6 x 10^-6 F

At resonance, the angular frequency is given by

[tex]\omega _{0}=\frac{1}{\sqrt{LC}}[/tex]

[tex]\omega _{0}=\frac{1}{\sqrt{1.05\times 2.6\times 10^{-6}}}=605.23 rad/s[/tex]

XL = ω0 x L = 605.23 x 1.05 = 635.5 ohm

Xc = 1 / ω0 C = 1 / (605.23 x 2.6 x 10^-6) = 635.5 ohm

In case of resonance, the impedance is equal to teh resistance of circuit.

Z = R = 21.1 ohm

I0 = E0 / Z = 90 / 21.1 = 4.265 A

Voltage across inductor

VL = I0 x XL

VL = 4.265 x 635.5 = 2710.66 V

To find the amplitude of the voltage across the inductor at resonance in an AC RLC circuit, we can calculate the resonant frequency using the formula fo = 1/(2*π√(LC)). Once we have the resonant frequency, we can find the amplitude of the voltage across the inductor using the formula Vl = I * XL, where XL = 2πfL.

To find the amplitude of the voltage across the inductor at resonance, we need to calculate the resonant frequency using the formula fo = 1/(2*π√(LC)).

Substituting the given values, fo = 1/(2 * 3.1416 * √(1.05 * 2.6 * 10^-6)) Hz.

Once we have the resonant frequency, we can find the amplitude of the voltage across the inductor using the formula Vl = I * XL, where XL = 2πfL.

Substituting the values and the calculated resonant frequency, we can find the amplitude of the voltage across the inductor.

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Final answer:

To find the nozzle's inside diameter, use the principle of continuity. Set up the equation A1V1 = A2V2 and solve for r2. Finally, calculate the nozzle's inside diameter.

Explanation:

To find the nozzle's inside diameter, we can use the principle of continuity, which states that the flow rate of a fluid is constant throughout a pipe or nozzle.

The formula for continuity is A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity.

In this case, we can set up the equation as (pi * r12) * 2.17 = (pi * r22) * 14.5, where r1 and r2 are the radii of the hose and nozzle respectively.

By rearranging the equation and solving for r2, we find that r2 = sqrt((r12 * 2.17) / 14.5).

Since the radius is half of the diameter, the nozzle's inside diameter can be calculated as 2 * r2 in cm.

Answer:

[tex]r=2.4m[/tex]

Explanation:

We have to use the centripetal force  equation

[tex]Fc=\frac{mv^{2} }{r}[/tex]

we  need the radious so we have to isolate "r" and we get

[tex]r=\frac{mv^{2} }{Fc}[/tex]

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

[tex]r=\frac{65*4.1^{2} }{455}[/tex]

[tex]r=2.4m[/tex]

The radius of the amusement park chamber is 2.4m

Final answer:

The work done by the force of push is 12 J. The work done by the force of friction is 7.08 J. The work done on the book by the force of gravity is zero.

Explanation:

To find the work done by the force of push, we can use the formula:

Work = Force x Distance

In this case, the force of push is 10 N and the distance is 1.2 m. So the work done by the force of push is:

Work = 10 N x 1.2 m = 12 J

To find the work done by the force of friction, we can use the same formula. In this case, the force of friction is 5.9 N and the distance is also 1.2 m. So the work done by the force of friction is:

Work = 5.9 N x 1.2 m = 7.08 J

The work done on the book by the force of gravity is zero since the book is being moved horizontally and gravity acts vertically.

Answer:

equal

Explanation:

According to the Newton's second law of motion, the force acting on the body is directly proportional to the rate of change of momentum.

If the force is same, then the change in momentum is also same for a given time.

The change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration.

The given data in the problem is;

F is the net force=200 N

m is the mass of boulder=100 kg

M is the mass of pebble=100 g

The force acting on the body is directly proportional to the rate of change in momentum, according to Newton's second law of motion.

For a given time, if the force is the same, the change in momentum is likewise the same.

Hence the change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.

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Answer:

5.95 nm to 33.6 nm

Explanation:

Energy of a single photon can be written as:

[tex]E = \frac{hc}{\lambda}[/tex]

where, h is the Planck's constant, c is the speed of light and λ is the wavelength of light.

Consider the lowest energy of an electron that can break a DNA = 3.45 eV

1 eV = 1.6 ×10⁻¹⁹ J

⇒3.45 eV =  5.52×10⁻¹⁹ J

[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{5.52 \times 10^{-19} J} = 3.60\times 10^{-7} m = 360 nm[/tex]

Consider the highest energy of an electron that can break a DNA = 20.9 eV

1 eV = 1.6 ×10⁻¹⁹ J

⇒20.9 eV =  33.4×10⁻¹⁹ J

[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{33.4 \times 10^{-19} J} = 0.595\times 10^{-7} m = 59.5 nm[/tex]

The wavelength range of light that can cause DNA breaks is approximately 59.4 nm to 360 nm. This corresponds to ultraviolet light and part of the visible spectrum.

To find the range of light wavelengths that can cause DNA breaks, we need to convert the given energy range of 3.45 eV to 20.9 eV into wavelengths.

The energy of a photon (E) is related to its wavelength (λ) by the equation:

E = hc/λ

where h is Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light .(3.00 × 10⁸m/s), and λ is the wavelength in meters.

First, convert the energy from electron volts (eV) to Joules (J):

So, the energy range in Joules is:

Next, use the energy-wavelength relation to find the wavelengths:

Thus, the range of wavelengths that can cause DNA breaks is approximately 59.4 nm to 360 nm, corresponding to the ultraviolet (UV) and part of the visible spectrum.

Answer:

14 x 10⁻⁶ J

1377 x 10⁻⁶ J

Explanation:

C = Capacitance of the capacitor = 1 x 10⁻⁶ F

ΔV = Original potential difference across the plates = 6.0 Volts

U₀ = Original energy stored in the capacitor

Original energy stored in the capacitor is given as

U₀ = (0.5) C ΔV²                                eq-1

ΔV' = Potential difference across the plates after increase = 8.0 Volts

U'₀ = New energy stored in the capacitor

New energy stored in the capacitor is given as

U'₀ = (0.5) C ΔV'²                                eq-2

U = Additional energy stored

Additional energy stored by the capacitor is given as

U = U'₀ - U₀

U = (0.5) C ΔV'² - (0.5) C ΔV²

U = (0.5) (1 x 10⁻⁶) (8)² - (0.5) (1 x 10⁻⁶) (6)²

U = 14 x 10⁻⁶ J

[tex]k_{final}[/tex] = final dielectric constant = 76.5

[tex]k_{initial}[/tex] = initial dielectric constant = 1

Energy stored in the capacitor is directly proportional to the dielectric constant, hence increase in the energy is given as

[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]

Original energy stored in the capacitor is given as

U₀ = (0.5) C ΔV²   = (0.5) (1 x 10⁻⁶) (6)² = 18 x 10⁻⁶ J

[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]

[tex]U_{inc} = (76.5 - 1)(18\times 10^{-6})[/tex]

[tex]U_{inc} = 1377\times 10^{-6})[/tex]

The amount of charge stored on the capacitor plates increases by a factor of [tex]$76.5}$[/tex].

The energy stored in a capacitor is given by the formula:

[tex]\[ E = \frac{1}{2} C V^2 \][/tex]

where [tex]$E$[/tex] is the energy, [tex]$C$[/tex] is the capacitance, and [tex]$V$[/tex] is the potential difference.

Initially, the energy stored in the capacitor with a potential difference of 6.0 V is:

[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (6.0)^2 \][/tex]

[tex]\[ E_1 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 36 \][/tex]

[tex]\[ E_1 = 18 \times 10^{-6} \text{ J} \][/tex]

[tex]\[ E_1 = 18 \text{ µJ} \][/tex] µJ

After increasing the potential difference to 8.0 V, the energy stored in the capacitor is:

[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times (8.0)^2 \][/tex]

[tex]\[ E_2 = \frac{1}{2} \times 1.0 \times 10^{-6} \times 64 \][/tex]

[tex]\[ E_2 = 32 \times 10^{-6} \text{ J} \][/tex]

[tex]\[ E_2 = 32 \text{ µJ} \][/tex] µJ

The additional energy stored is the difference between [tex]$E_2$[/tex] and [tex]$E_1$[/tex]:

[tex]\[ \Delta E = E_2 - E_1 \][/tex]

[tex]\[ \Delta E = 32 - 18 \][/tex]

[tex]\[ \Delta E = 14 \text{ µJ} \][/tex] µJ

Now, let's consider the effect of changing the dielectric constant from 1 to 76.5. The capacitance of a capacitor with a dielectric material is given by:

[tex]\[ C = \frac{k \epsilon_0 A}{d} \][/tex]

Since the dielectric constant [tex]$k$[/tex] is the only quantity changing, the new capacitance [tex]$C'$[/tex] is:

[tex]\[ C' = kC \][/tex]

where [tex]$C$[/tex] is the original capacitance with [tex]$k = 1$[/tex].

Therefore, [tex]$C' = 76.5 \times 1.0 \times 10^{-6} \text{ F}$[/tex].

The charge stored on a capacitor is given by:

[tex]\[ Q = CV \][/tex]

With the new capacitance and the same potential difference of 8.0 V, the new charge [tex]$Q'$[/tex] is:

[tex]\[ Q' = C'V \][/tex]

[tex]\[ Q' = 76.5 \times 1.0 \times 10^{-6} \times 8.0 \][/tex]

[tex]\[ Q' = 612 \times 10^{-6} \text{ C} \][/tex]

The original charge [tex]$Q$[/tex] with [tex]$k = 1$[/tex] and [tex]$V = 8.0 \text{ V}$[/tex] was:

[tex]\[ Q = C \times V \][/tex]

[tex]\[ Q = 1.0 \times 10^{-6} \times 8.0 \][/tex]

[tex]\[ Q = 8 \times 10^{-6} \text{ C} \][/tex]

The ratio of the new charge to the original charge is:

[tex]\[ \frac{Q'}{Q} = \frac{612 \times 10^{-6}}{8 \times 10^{-6}} \][/tex]

[tex]\[ \frac{Q'}{Q} = 76.5 \][/tex]

Answer:

Centripetal force, F = 18486.75 N

Explanation:

It is given that,

Mass of the sensor, m = 6 kg

It is attached at the end of 100 m long wind turbine, d = 100 m

So, the radius of the wind turbine, r = 50 m

Angular velocity, [tex]\omega=1.25\ rev/s=7.85\ rad/s[/tex]

The centripetal force is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

Since, [tex]v=r\omega[/tex]

[tex]F=mr\omega^2[/tex]

[tex]F=6\ kg\times 50\ m\times (7.85\ rad/s)^2[/tex]

F = 18486.75 N

So, the centripetal force is 18486.75 N. Hence, this is the required solution.

Answer:

104

Explanation:

U = Energy stored in the solenoid = 6.00 μJ = 6.00 x 10⁻⁶ J

i = current flowing through the solenoid = 0.4 A

L = inductance of the solenoid

Energy stored in the solenoid is given as

U = (0.5) L i²

6.00 x 10⁻⁶ = (0.5) L (0.4)²

L = 75 x 10⁻⁶

Inductance is given as

l = length of the solenoid = 0.7 m

N = number of turns

r = radius = 5.00 cm = 0.05 m

Area of cross-section is given as

A = πr²

A = (3.14) (0.05)²

A = 0.00785 m²

Inductance is given as

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

[tex]75\times 10^{-6}=\frac{(12.56\times 10^{-7})N^{2}(0.00785)}{0.7}[/tex]

N = 73

Winding density is given as

density = n = [tex]\frac{N}{l}[/tex]

n = [tex]\frac{73}{0.7}[/tex]

n = 104

Answer:

The speed of the wire is 5 m/s.

Explanation:

Given that,

Length = 20 cm

Magnetic field = 0.1 T

Current = 10 mA

Resistance [tex]R= 10\Omega[/tex]

We need to calculate the  speed of the wire

Using formula of emf

[tex]\epsilon=Bvl[/tex]

Using formula of current

[tex]I=\dfrac{\epsilon}{R}[/tex]

Put the value of [tex]\epsilon[/tex] into the formula of current

[tex]I=\dfrac{Bvl}{R}[/tex]

[tex]v=\dfrac{IR}{Bl}[/tex]

[tex]v=\dfrac{10\times10^{-3}\times10}{0.1\times20\times10^{-2}}[/tex]

[tex]v= 5\ m/s[/tex]

Hence, The speed of the wire is 5 m/s.

Answer:

Magnitude of the net force acting on the kayak = 39.61 N

Explanation:

Considering motion of kayak:-

Initial velocity, u =  0 m/s

Distance , s = 0.40 m

Final velocity, v = 0.65 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0.65² = 0² + 2 x a x 0.4

    a = 0.53 m/s²

We have force, F  = ma

Mass, m = 75 kg

    F  = ma = 75 x 0.53 = 39.61 N

Magnitude of the net force acting on the kayak = 39.61 N

Answer:

Moe: 6687.5 J, Larry: -2812.5 J, Curly: 0 J

Explanation:

The work done by each clown is given by:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force applied

d is the displacement of the box

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

Let's apply the formula to each clown:

- Moe: F = 535 N, d = 12.5 m, [tex]\theta=0^{\circ}[/tex] (because Moe pushes to the right, and the box also moves to the right)

[tex]W=(535)(12.5)cos 0^{\circ}=6687.5 J[/tex]

- Larry: F = 225 N, d = 12.5 m, [tex]\theta=180^{\circ}[/tex] (because Larry pushes to the left, while the box moves to the right)

[tex]W=(225)(12.5)cos 180^{\circ}=-2812.5 J[/tex]

- Curly: F = 705 N, d = 12.5 m, [tex]\theta=90^{\circ}[/tex] (because Curly pushes downward, while the box moves to the right)

[tex]W=(705)(12.5)cos 90^{\circ}=0[/tex]

The work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

Work done is the force applied on a body to move it over a distance. Work done to move a body by the application of force can be given as,

[tex]W=Fd\cos\theta[/tex]

Here (F) is the magnitude of force, [tex]\theta[/tex] is the angle of force applied and (d) is the distance traveled.

The mass of the crate is 345 kg. The crate moved by the clowns is 12.5 m to the right across a smooth floor.

As, the Moe applies the force to the crate on the right side and the crate move in the direction of force. Here, the angle made is equal to the 0 degrees.

Therefore, the work done by the Moe is,

[tex]W=535\times12.5\times\cos (0)\\W=6687.5\rm J[/tex]

thus, the work done by the Moe is 6687.5 J.

As, the Larry applies the force to the crate on the left side and the crate move in the opposite direction of force. Here, the angle made is equal to the 180 degrees.

Therefore, the work done by the Larry is,

[tex]W=225\times12.5\times\cos (180)\\W=-2812.5\rm J[/tex]

thus, the work done by the Larry is -2812.5 J.

As, the Curly applies the force to the crate on the straight down and the crate move in the perpendicular direction of force. Here, the angle made is equal to the 90 degrees.

Therefore, the work done by the Curly is,

[tex]W=705\times12.5\times\cos (90)\\W=0\rm J[/tex]

thus, the work done by the Curly is 0 J.

Thus, the work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

Learn more about the work done here;

https://brainly.com/question/25573309

Answer:

Answer to the question:

Explanation:

Exothermic Reaction

It is called an exothermic reaction to any chemical reaction that releases energy, either as light or heat,  or what is the same: with a negative variation of enthalpy; that is to say: ΔH < 0. Therefore it is understood that exothermic reactions release energy.

Endothermic Reaction

It is called an endothermic reaction to any chemical reaction that absorbs energy, usually in the form of heat.

If we talk about enthalpy (H), an endothermic reaction is one that has a variation of enthalpy ΔH> 0. That is, the energy possessed by the products is greater than that of the reagents.

Answer:

(a) 0.816 s

(b) 0.816 m

(c) 5.656 m

Explanation:

u = 8 m/s, theta = 30 degree,

(a) Use the formula for time of flight

T = 2 u Sin theta / g

T = ( 2 x 8 x Sin 30 ) / 9.8 = 0.816 s

(b) Use the formula for maximum height

H = u^2 Sin^2theta / 2 g

H = ( 8 x 8 x Sin^2 30) / ( 2 x 9.8)

H = 0.816 m

(c) Use the formula for horizontal range

R = u^2 Sin 2 theta / g

R = ( 8 x 8 x Sin 60) / 9.8

R = 5.656 m

Answer:

[tex]F=55146000 N[/tex]

Explanation:

Given:

Pressure on the building = 170 N/m²

Pressure function P(y) = 170 + 2y

Consider a small height of the building as dy

the small force (dF) exerted on the height dy will be given as

dF = P(y)×(Area of the building face)

or

dF = (170 + 2y)(6565)dy

or

dF = 170×6565×dy + 2y×6565×dy

dF = 1116050dy + 13130ydy

integerating the function for the whole height of 40 m

[tex]\int\limits^{40}_0 {F} = \int _0^{40}\:\left(\:1116050dy\:+\:13130ydy\right)[/tex]

[tex]F = \int _0^{40}1116050dy+\int _0^{40}13130ydy[/tex]

[tex]F=\left[1116050y\right]^{40}_0+13130\left[\frac{y^2}{2}\right]^{40}_0[/tex]

[tex]F=44642000+10504000[/tex]

[tex]F=55146000 N[/tex] ; Total force exerted on the face of the building

To calculate the total force on a building due to wind pressure, integrate the pressure function over the building's height and multiply by the building's width. This calculation is vital for determining the building's required resistance to wind forces for safe and stable design.

The question asks to calculate the total force exerted on a large building by a wind blowing against one of its faces. The wind pressure increases with height as described by the function P(y)=170+2y, where y is the height in meters above the ground. To find the total force, we need to integrate this pressure function over the height of the building and multiply by the width of the building's face.

First, we can write the integral of the pressure over the height of the building:

Integrate [P(y) dy] from y=0 to y=40

= Integrate [(170+2y) dy] from y=0 to y=40

Performing this integration will give us the total pressure exerted over the entire height of the building. After integrating, we multiply the result by the width of the building (6565 m) to obtain the total force.

Using the integration results, the total force, F, can be expressed as:

F = (Integral result) × Width of the building

This force calculation is critical for architects and engineers as it provides an estimate of the resistance that the building must be designed to withstand due to wind forces, ensuring structural stability and safety.

Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)

Explanation:

The sign of charge Q2 is positive, and its magnitude can be calculated by setting the gravitational force equal to the electric force due to Q1 and solving for Q2.

To determine the sign and magnitude of the second charge Q2, we use the concept that the net force acting on it must be zero. The forces that act on Q2 include the gravitational force and the electric force due to Q1. Since Q1 is positive and the net force is zero, Q2 must also be positive for the electric force (Coulomb force) to balance the gravitational pull downwards.

The gravitational force can be calculated using Newtons's law of gravitation F = mg, where g is the acceleration due to gravity (9.8 m/s2). The electric force is given by Coulomb's law F = k|Q1Q2|/d2, where k is Coulomb's constant (8.99 x 109 Nm2/C2).

Setting the magnitude of the gravitational force equal to the magnitude of the electric force gives the equation mg = k|Q1Q2|/d2. Solving for Q2, and converting the mass of Q2 from micrograms to kilograms, we calculate the magnitude of charge Q2.