The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\times 10^{-5}~\text{N}\cdot\text{m}^2/\text{C}3×10 ​−5 ​​ N⋅m ​2 ​​ /C when the area is parallel to the sheet of charge. Find the charge density on the sheet.

Answer:

a) [tex]v_{max}=0.4\ m.s^{-1}[/tex]

b) [tex]a_{max}=1.6\ m.s^{-2}[/tex]

c) [tex]v_x=0.32\ m.s^{-1}[/tex]

d) [tex]a_x=0.96\ m.s^{-1}[/tex]

e) [tex]\Delta t=0.232\ s[/tex]

Explanation:

Given:

mass of the object attached to the spring, [tex]m=0.5\ kg[/tex]

spring constant of the given spring, [tex]k=8\ N.m^{-1}[/tex]

amplitude of vibration, [tex]A=0.1\ m[/tex]

a)

Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.

Max. spring potential energy:

[tex]PE_s=\frac{1}{2} .k.A^2[/tex]

[tex]PE_s=0.5\times 8\times 0.1^2[/tex]

[tex]PE_s=0.04\ J[/tex]

When this whole spring potential is converted into kinetic energy:

[tex]KE_{max}=0.04\ J[/tex]

[tex]\frac{1}{2}.m.v_{max}^2=0.04[/tex]

[tex]0.5\times 0.5\times v_{max}^2=0.04[/tex]

[tex]v_{max}=0.4\ m.s^{-1}[/tex]

b)

Max. Force of spring on the mass:

[tex]F_{max}=k.A[/tex]

[tex]F_{max}=8\times 0.1[/tex]

[tex]F_{max}=0.8\ N[/tex]

Now acceleration:

[tex]a_{max}=\frac{F_{max}}{m}[/tex]

[tex]a_{max}=\frac{0.8}{0.5}[/tex]

[tex]a_{max}=1.6\ m.s^{-2}[/tex]

c)

Kinetic energy when the displacement is, [tex]\Delta x=0.06\ m[/tex]:

[tex]KE_x=PE_s-PE_x[/tex]

[tex]\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2[/tex]

[tex]\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2[/tex]

[tex]v_x=0.32\ m.s^{-1}[/tex]

d)

Spring force on the mass at the given position, [tex]\Delta x=0.06\ m[/tex]:

[tex]F=k.\Delta x[/tex]

[tex]F=8\times 0.06[/tex]

[tex]F=0.48\ N[/tex]

therefore acceleration:

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

[tex]a_x=\frac{0.48}{0.5}[/tex]

[tex]a_x=0.96\ m.s^{-1}[/tex]

e)

Frequency of oscillation:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{8}{0.5} }[/tex]

[tex]\omega=4\ rad.s^{-1}[/tex]

So the wave equation is:

[tex]x=A.\sin\ (\omega.t)[/tex]

where x = position of the oscillating mass

put x=0

[tex]0=0.1\times \sin\ (4t)[/tex]

[tex]t=0\ s[/tex]

Now put x=0.08

[tex]0.08=0.1\times \sin\ (4t)[/tex]

[tex]t=0.232\ s[/tex]

So, the time taken in going from point x = 0 cm to x = 8 cm is:

[tex]\Delta t=0.232\ s[/tex]

Final answer:

The problem involves calculating the maximum speed, maximum acceleration, speed and acceleration at a certain distance from equilibrium, and the time interval for an object in simple harmonic motion. Using formulas for SHM including maximum speed (v_max = ωA), maximum acceleration (a_max = ω^2A), and expressions for speed and acceleration at a given position, one can determine these values for the object on the spring.

Explanation:

The student has asked us to calculate various properties of an object undergoing simple harmonic motion (SHM) when attached to a spring with a known force constant and amplitude. To answer this question, one needs to use equations that describe SHM.

Maximum Speed (v_max) Calculation:

The maximum speed (v_max) of an object in SHM occurs when it passes through the equilibrium point and can be calculated using the formula v_max = ωA, where ω is the angular frequency (ω = sqrt(k/m)) and A is the amplitude of the motion.

Maximum Acceleration (a_max) Calculation:

The maximum acceleration (a_max) occurs at the maximum displacement and is given by a_max = ω^2A.

Speed at 6 cm from Equilibrium:

To find the speed at a certain position x, we use the formula v = ω sqrt(A^2 - x^2).

Acceleration at 6 cm from Equilibrium:

Acceleration at any position x is a = -ω^2x.

Time Interval to Move from 0 to 8 cm:

The time interval to move from x = 0 to a certain position x can be found using the formula for time in SHM as a function of position.

Answer:

0.34m

Explanation:

Final answer:

The wavelength of a radio signal received at a frequency of 880.65 MHz is approximately 0.34 meters, using the formula λ = c / f, where c is the speed of light.

Explanation:

If you receive a call at a frequency of 880.65 MHz, to calculate the wavelength in meters, you can use the formula λ = c / f, where λ is the wavelength in meters, c is the speed of light (approximately 3.00 × 108 m/s), and f is the frequency in hertz (Hz). Since the question provides the frequency in megahertz (MHz), we first convert it to hertz by multiplying by 106.

The frequency is 880.65 MHz, which is equal to 880.65 × 106 Hz. Thus, the wavelength λ = 3.00 × 108 m/s / (880.65 × 106 Hz).

Calculating this gives a wavelength of approximately 0.34 meters.

Answer:

Explanation:

Given

Length of cable [tex]L=90\ m[/tex]

mass of cable [tex]m=100\ kg[/tex]

Tension in the cable [tex]T=25,000\ N[/tex]

The lowest frequency observed is given by

[tex]\nu =\frac{1}{2L}\cdot \sqrt{\frac{T}{\mu }}[/tex]

where [tex]\nu =frequency[/tex]

[tex]\mu =mass\ density(kg/m)[/tex]

[tex]\mu =\frac{100}{90}=\frac{10}{9}\ kg/m[/tex]

[tex]\nu =\frac{1}{2\times 90}\cdot \sqrt{\frac{25,000}{\frac{10}{9}}}[/tex]

[tex]\nu =\frac{5\times 10\times 3}{2\times 90}[/tex]

[tex]\nu =0.833\ Hz[/tex]                            

Good conductors of electricity have larger conductivity values than insulators.

A material that obeys Ohm's law reasonably well is called an ohmic conductor or a linear conductor.

The resistance of a conductor is proportional to the conductivity of the material of which the conductor is composed.

Answer: Options 1, 2 and 4.

Explanation:

In physics and electrical engineering, a conductor is an article or kind of material that permits the progression of charge in at least one headings. Materials made of metal are basic electrical conduits.

Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them. Most atoms hold on to their electrons tightly and are insulators.

Answer:[tex]N=8.28\ rpm[/tex]

Explanation:

Given

Diameter of wheel [tex]d=26\ m[/tex]

Person is feeling Weightlessness i.e. Net force on the person is equivalent to its weight

At top point weight is equal to Centripetal force

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

where v=velocity of wheel

thus

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

[tex]v=\sqrt{gR}[/tex]

[tex]v=\sqrt{9.8\times 13}[/tex]

[tex]v=11.28\ m/s[/tex]

[tex]v=\frac{\pi d\cdot N}{60}[/tex]

[tex]11.28=\frac{\pi \cdot 26\cdot N}{60}[/tex]

[tex]N=8.28\ rpm[/tex]

Answer:

Explanation:

Let m be the mass of passenger.

diameter of wheel, d = 26 m

radius of wheel, r = half of diameter = 13 m

Let ω be the angular velocity of the Ferris wheel.

A the passengers becomes weightless, so the centripetal force acting on the passengers is balanced by the weight of passengers.

mg = m r ω²

9.8 = 12 x ω²

ω = 0.9 rad/s

Let f be the frequency

ω = 2 π f

0.9 = 2 x 3.14 x f

f = 0.143 revolutions per second

Number of revolutions per minute = 0.143 x 60

                                                         = 8.6 revolutions per minute

Answer:

297.8 m

Explanation:

We are given that

Acceleration=[tex]a=1.65m/s^2[/tex]

Time,t=19 s

We have to find the distance covered by the train when it reach its top speed if starting from rest.

Initial speed=u=0

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

Using the formula

[tex]s=0(19)+\frac{1}{2}\times 1.65(19)^2[/tex]

[tex]s=297.8 m[/tex]

Hence, the train covered 297.8 m when the train go to reach its top speed if starting from rest.

Answer:

4.has gained two electrons

Explanation:

There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.

Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2

Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6

This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.

Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.

Final answer:

In MgS, the sulfide ion has gained two electrons, resulting in a charge of -2, represented as S²-. Magnesium loses two electrons to form Mg²+, balancing the electron transfer to create a stable ionic compound.

Explanation:

In the compound MgS, the sulfide ion has gained two electrons to achieve a stable electron configuration. When sulfur (S), which has an atomic number of 16, gains two electrons, it results in an ion with 18 electrons and 16 protons. This gives the ion a charge of -2, since there are two more negative electrons than positive protons. Therefore, the sulfide ion is represented as S²-. The magnesium atom loses its two valence electrons to become a Mg²+ cation, as magnesium is in Group 2A of the periodic table and tends to lose two electrons to achieve a noble gas electron configuration. This electron transfer process is balanced, meaning the number of electrons lost by magnesium is equal to the number of electrons gained by sulfur, forming a stable ionic compound.

Answer:

Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C

Explanation:

Given,A = side of the triangle. q = 1 C

The center of a triangle is the centroid of the triangle.

The distance between centroid to any vertices of a equilateral triangle is

[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle

[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]

[tex]=\frac{1}{\sqrt{3} } A[/tex]

Electric field= [tex]\frac{Kq}{d^2}[/tex]       K=8.99×10⁹ Nm²/C², q=charge  and d = distance

Therefore the magnitude of the net electric filed at the center of the triangle is

=2[tex]\frac{Kq}{d^2}[/tex]         [both charge are at same distance from the centroid]

=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]

[tex]=\frac{6K}{A^2}[/tex] N /C      [K=8.99×10⁹ Nm²/C²]

Answer:

1230 km  

Explanation:

From the time delay, we can write:

[tex]t_{s} -t_{p} =[/tex]Δt

Knowing that t =[tex]\frac{d}{v}[/tex]  we rewrite the formula as  

[tex]\frac{d}{vs} -\frac{d}{vp} =d(\frac{1}{vp}-\frac{1}{vs} )[/tex]=Δt

From this we can find the distance to be  

[tex]d=\frac{v_{s} v_{p} }{v_{p}-v_{s} }[/tex]*Δt

   = 1230 km  

To find the distance to an earthquake's epicenter, the difference in arrival times of P-waves and S-waves is used. With a 2.3-minute delay, and known velocities for both types of waves, the earthquake's epicenter is determined to be 483 kilometers away.

To calculate the distance to the epicenter of an earthquake, geologists use the difference in arrival times of seismic waves: the faster longitudinal waves (P-waves) and the slower transverse waves (S-waves). Given that P-waves travel at 8000 m/s and S-waves at 4500 m/s, and the S waves arrive 2.3 minutes after the P waves, we can determine the distance to the earthquake's epicenter.

First we convert the time delay from minutes to seconds: 2.3 minutes is 2.3 × 60 seconds = 138 seconds. The difference in distance covered by the two types of waves in this time can be calculated by multiplying the speed of each wave by the time delay:

The difference in distances gives us the distance to the epicenter:

Distance to epicenter = Distance covered by P-waves - Distance covered by S-waves

                        = (8000 m/s × 138 s)  - (4500 m/s × 138 s)

                        = 1104000 m - 621000 m

                        = 483000 m or 483 km

Hence, the earthquake was 483 kilometers away from the seismograph recording station.

Final answer:

To calculate the new velocity of the boat after being accelerated by the current for 5 seconds, one must use the Pythagorean theorem and arctangent function to find the magnitude and direction of the resultant velocity, resulting in approximately 5.04 m/s at 48.1 degrees from the original motion.

Explanation:

The student's question is about calculating the resultant velocity of a boat being accelerated by a river current perpendicular to its initial direction of motion. With an initial velocity of 3.35 m/s and a river current accelerating the boat at 0.750 m/s2, we need to find the new velocity after 5 seconds.

First, calculate the velocity increase caused by the acceleration of the current: velocity increase = acceleration × time, which gives us 0.750 m/s2 × 5 s = 3.75 m/s. This increase is perpendicular to the initial velocity.

Now, we determine the resultant velocity using the Pythagorean theorem since the velocities are perpendicular. The magnitude of the resultant velocity, Vtotal, is given by √(3.35 m/s)2 + (3.75 m/s)2, which equals to approximately 5.04 m/s. To find the direction, we use the arctangent function: θ = tan-1(3.75 m/s / 3.35 m/s), which results in a direction of approximately 48.1 degrees relative to the original direction of motion. Therefore, the boat will be moving with a velocity of approximately 5.04 m/s at a direction of approximately 48.1 degrees from its original direction, due to the current.

Answer:

7. The force is zero

Explanation:

The force is zero when your velocity is parallel to the magnetic field

Answer:

doppler shift's formula for source and receiver moving away from each other:

λ'=λ°√(1+β/1-β)

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

λ'=λ°√(1+β/1-β)

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

β=0.17

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

v=v₀+at

where v₀=0

so v=at

as we want to calculate t so:-

t=v/a                                                v=0.17c      ,c=3x10⁸           ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

t=1.73x10⁶

Answer:

v_b = 10.628 m /s (13.605 degrees east of south)

Explanation:

Given:

- Velocity of mia v_m = + 6 j m/s

- Velocity of ball wrt mia v_bm = 5.0 m/s     30 degree due east of south

Find:

What are the magnitude and direction of the velocity of the ball relative to the ground? v_b

Solution:

- The relation of velocity in two different frame is given:

                             v_b  - v_m = v_bm

- Components along the direction of v_b,m:

                             v_b*cos(Q) - v_m*cos(30) = 5

                             v_b*cos(Q) = 5 + 6 sqrt(3) / 2

                             v_b*cos(Q) = 5 + 3sqrt(3)  

- Components orthogonal the direction of v_b,m:

                             -v_m*sin(30) = v_b*sin(Q)

                             -6*0.5 = v_b*sin(Q)

                              -3 = v_b*sin(Q)

- Divide two equations:

                               tan(Q) = - 3 / 5 + 3sqrt(3)

                               Q = arctan(- 3 / 5 + 3sqrt(3)

                               Q = -16.395 degrees

                               v_b =  -3 / sin(-16.395)

                               v_b = 10.63 m/s

The magnitude of the soccer ball's velocity relative to the ground is 2.89 m/s, and its direction is approximately 56.31° north of east. This result is found by breaking down vector components and applying vector addition.

You're asking about the velocity of the ball relative to the ground when soccer players Mia and Alice are running, and Alice passes the ball to Mia. To solve this, we'll use vector addition. Mia is running due north at 6.00 m/s, and the velocity of the ball relative to Mia is 5.00 m/s at 30° east of south. To find the velocity of the ball relative to the ground, we imagine two vectors: Mia's velocity vector (northward) and the ball's velocity vector relative to Mia. The latter will have an east and a south component due to the 30° angle.

To find the south component of the ball's relative velocity, we use cosine because it's adjacent to the 30° angle:
5.00 m/s * cos(30°) = 4.33 m/s. The east component is found using sine:
5.00 m/s * sin(30°) = 2.50 m/s.

Since Mia is running north, to find the actual velocity of the ball to the south, we subtract Mia's velocity from the south component:
4.33 m/s - 6.00 m/s = -1.67 m/s, where the negative indicates that the ball's actual movement is to the north.

Now, using the Pythagorean theorem for the ball's velocity relative to the ground, we find the magnitude:
(2.50 m/s)² + (-1.67 m/s)² = √(6.25 + 2.7889)
√8.3389 m²/s² = 2.89 m/s.

To find the direction, we use the tangent function, since we have opposite (east component) and adjacent (north component) sides of the right triangle:
an(θ) = 2.50 / 1.67; θ = tan(⁻¹)(2.50 / 1.67);
θ ≈ 56.31° north of east, which is the direction of the ball's velocity relative to the ground.

Answer:

d) Gravity, a normal force, and kinetic friction

Explanation:

Answer:

Explanation:

Given

Man walks 17.5 m straight to west 17.5 m

So position vector is given by

[tex]\vec{r_1}=-17.5\hat{i}[/tex]

Now he walks 22 m North

so position vector is

[tex]r_{21}=22\hat{j}[/tex]

Position of man from initial Position

[tex]\vec{r_{2}}=\vec{r_2}-\vec{r_1}[/tex]

[tex]\vec{r_{2}}=22\hat{j}-(-17.5\hat{i})[/tex]

[tex]\vec{r_{2}}=17.5\hat{i}+22\hat{j}[/tex]

So Magnitude of distance is given by

[tex]|\vec{r_{2}}|=\sqrt{17.5^2+22^2}[/tex]

[tex]|\vec{r_{2}}|=28.11\ m[/tex]  

To determine the distance from the starting point after walking 17.5 m west and 22.0 m north, use the Pythagorean theorem. Calculating this gives a distance of approximately 28.1 meters from the starting point.

Calculating Distance Using Pythagorean Theorem

To find how far you are from your starting point after walking 17.5 m west and then 22.0 m north, we can use the Pythagorean theorem. This is because your path forms a right triangle with the two legs being 17.5 m and 22.0 m.

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the distance from your starting point) is equal to the sum of the squares of the other two sides:

a² + b² = c²

Substituting these values into the Pythagorean theorem gives:

17.52 + 22.02 = c²

Calculating the squares:

306.25 + 484.00 = c²

Adding them together:

790.25 = c²

Taking the square root of both sides:

c ≈ 28.1 m

Therefore, you are approximately 28.1 meters from your starting point.

Final answer:

Spectroscopy is a technique used in physics to study the absorption, emission, or scattering of electromagnetic radiation by atoms or molecules.

Explanation:

Spectroscopy is a technique used in physics to study the absorption, emission, or scattering of electromagnetic radiation by atoms or molecules. It is commonly used to analyze and understand the composition, temperature, and motion of celestial objects like stars and galaxies. By observing the patterns of light emitted or absorbed by these objects, astronomers can determine the elements present and gain insights into their physical properties.

Answer:

[tex]E=0[/tex] at r < R;

[tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex] at 2R > r > R;

[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex] at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

[tex]\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}[/tex] (integral over a closed surface)

where,

[tex]E[/tex] = Electric field

[tex]Q_{enclosed}[/tex] = charged enclosed within the closed surface

[tex]\epsilon[/tex] = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

[tex]Q_{enclosed}[/tex] = 0 and hence [tex]E[/tex] = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

[tex]Q_{enclosed}[/tex] = Q,

therefore,

[tex]E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}[/tex]      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to [tex]4\pi r^{2}[/tex])

or, [tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex]

at r >= 2R

[tex]Q_{enclosed}[/tex] = 2Q

Hence, by similar calculations, we get,

[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex]

The electric field inside the solid conducting sphere is zero. For r between R and 2R, the electric field can be calculated using Gauss's law.

The electric field inside a solid conducting sphere is zero. Thus, the magnitude of the electric field in the region 0<r<R is zero.

In the region R<r<2R, the charge on the insulating shell induces an equal and opposite charge on the inner surface of the conducting sphere. Therefore, the magnitude of the electric field in this region can be found using Gauss's Law.

The electric field magnitude is given by:

E = Q / (4πε0 r2)

where Q is the charge on the insulating shell, ε0 is the vacuum permittivity, and r is the distance from the center of the sphere.

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Answer:

F_net = - 0.365 N (Down-ward direction)

Explanation:

Given:

- Value of r = 0.75 m

- Charges on x axis are -2*q

- Charge +q on origin

- Charge on - y axis is -2q

- Charge on + y axis is +q

- q = 5.00 * 10^-6 C

Find:

-What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls?

Solution:

- Force due to each of the two charges on x axis:

                                  F_x = k*(-2*q)*(+q) / r*^2

                                  r* = sqrt(2)*r

                                  F_x = -k*q^2 / r^2 (Down-wards)

- Force due to +q charge on origin:

                                  F_y = k*(+q)*(+q) / r^2

                                  F_y = + k*q^2 / r^2 (Up-wards)

- Force due to -2*q charge on y-axis:

                                  F_-2y = k*(-2*q)*(+q) / 4*r^2

                                  F_-2y = - k*q^2 / 2*r^2 (Downwards-wards)

- Total net Force on charge +q on + y-axis:

                                  2*F_x*sin(45) + F_y + F_-2y = F_net

                                 -sqrt(2)*k*q^2 / r^2 + k*q^2 / r^2 - k*q^2 / 2*r^2 = F_net

                                 (0.5-sqrt(2))*k*q^2 / r^2 = F_net

                                  F_net =  (0.5-sqrt(2))*(8.99*10^9)*(5*10^-6)^2 / 0.75^2

                                  F_net = - 0.365 N

Answer:

2.03 Ω

Explanation:

EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.

The formula of emf is given as,

E = I(R+r)............................ Equation 1

Where E = Emf, I = current, R = External resistance, r = internal resistance.

Make r the subject of the equation

r = (E-IR)/I........................ Equation 2

Note: From ohm's law, V = IR.

r = (E-V)/I........................ Equation 3

Where V = Terminal voltage

Given: E = 24 V, I = 3.9 A, V = 16.1 V.

Substitute into equation 3

r = (24-16.1)/3.9

r = 7.9/3.9

r = 2.03 Ω

Final answer:

To find the internal resistance of a battery, subtract the terminal voltage from the emf and divide by the current. Given an emf of 24.0 V, a terminal voltage of 16.1 V, and a current of 3.90 A, the internal resistance is about 2.03 Ω.

Explanation:

To determine the internal resistance of a battery, we need to understand a critical relationship between the battery's electromotive force (emf), its terminal voltage, the current flowing through the circuit, and its internal resistance. The formula required for finding the internal resistance is derived from Ohm's Law, and it takes into account the difference between the emf (the voltage the battery would supply in the absence of internal resistance) and the terminal voltage (the actual voltage the battery supplies when connected to a load).

Given are the following parameters:

emf (E) = 24.0 V

Terminal Voltage (V) = 16.1 V

Current (I) = 3.90 A

To calculate the internal resistance (r), we use the formula based on the definition of emf, which is:

E = V + Ir

By rearranging the formula to solve for the internal resistance, we get:

r = (E - V) / I

Substituting the given values:

r = (24.0 V - 16.1 V) / 3.90 A = 7.9 V / 3.90 A ≈ 2.03 Ω

Therefore, the internal resistance of the battery is approximately 2.03 Ω.

Answer:

v = 9.824 x 10³ m/s

Explanation:

given,

Linear momentum of photon,λ = 740 n m

for photon,

[tex]p=\dfrac{h}{\lambda}[/tex]

h is the planks constant

P is the momentum

[tex]p=\dfrac{6.626\times 10^{-34}}{740\times 10^{-10}}[/tex]

      p = 8.95 x 10⁻²⁷ kg.m/s

For electron

p = m v

mass of electron = 9.11 x 10⁻³¹ Kg

[tex]v = \dfrac{8.95\times 10^{-27}}{9.11\times 10^{-31}}[/tex]

v = 9.824 x 10³ m/s

hence, the velocity of electron is equal to   v = 9.824 x 10³ m/s

Answer:

s = 23.68 m

Explanation:

given,

acceleration of the sperm whale = 0.09 m/s²

initial speed of the whale, u = 0.7 m/s

final speed of the whale, v = 2.18 m/s

Distance traveled by the sperm whale = ?

Using equation of motion

 v² = u² + 2 a s

 2.18² = 0.7² + 2 x 0.09 x s

 0.18 s = 4.2624

 s = 23.68 m

Distance traveled by the whale is equal to s = 23.68 m

Answer:

s = 9.81t²

Explanation:

The equation of the vertical displacement of the unicorn at any time t, can be derived using the equation of linear motion

s = Vo.t + 0.5at²

where s is the displacement,

Vo is the initial velocity,

t the change in time from t = 0

a is the acceleration which for bodies falling under gravity without air resistance is 9.81 m/s²

So the equation simplifies to

s = 4.905t²

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

                                   V_f = sqrt(650.9861)

                                  V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

Answer: 1.6 m/s

Explanation:

The relationship between linear speed and angular speed for a rotational motion is

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

where

v is the linear speed

[tex]\omega[/tex] is the angular speed

r is the distance from the axis of rotation

For the vinyl record here, we have:

[tex]\omega=90 rev/min[/tex]

Keeping in mind that

[tex]1rev = 2\pi rad\\1 min =60 s[/tex]

We can convert it to rad/s:

[tex]\omega = 90 \cdot \frac{2\pi}{60}=9.4 rad/s[/tex]

The diameter of the disk is 34 cm, so the radius is

[tex]r=\frac{34}{2}=17 cm = 0.17 m[/tex]

Therefore, the linear velocity of  a point on the circumference is

[tex]v=(9.4)(0.17)=1.6 m/s[/tex]

The linear velocity of a point on the circumference of the disk is approximately 1.45 meters per second.

To find the linear velocity of a point on the circumference of the disk, we need to calculate the circumference of the disk and then determine how fast a point on the edge of the disk moves in one minute.

[tex]\[ v = \frac{C}{t} \][/tex]

First, we calculate the circumference  using the diameter of the disk. The radius is half of the diameter, so:

[tex]\[ r = \frac{diameter}{2} = \frac{34 \text{ cm}}{2} = 17 \text{ cm} \][/tex]

Now, the circumference  is given by:

[tex]\[ C = 2\pi r = 2\pi \times 17 \text{ cm} \][/tex]

Converting centimeters to meters (1 meter = 100 centimeters), we get:

[tex]\[ C = 2\pi \times 0.17 \text{ m} \][/tex]

Next, we know that the record makes 90 rotations in a minute, so the time \( t \) for one rotation is:

[tex]\[ t = \frac{60 \text{ seconds}}{90 \text{ rotations}} \][/tex]

Now we can calculate the linear velocity \( v \):

[tex]\[ v = \frac{C}{t} = \frac{2\pi \times 0.17 \text{ m}}{\frac{60}{90} \text{ seconds}} \] \[ v = \frac{2\pi \times 0.17 \text{ m}}{\frac{2}{3} \text{ seconds}} \] \[ v = \frac{2\pi \times 0.17 \text{ m} \times 3}{2} \] \[ v = \pi \times 0.17 \text{ m} \times 3 \] \[ v = \pi \times 0.51 \text{ m} \] Using the approximation \( \pi \approx 3.14159 \), we get: \[ v \approx 3.14159 \times 0.51 \text{ m/s} \] \[ v \approx 1.594 \text{ m/s} \][/tex]

Rounding to two decimal places, the linear velocity is approximately 1.45 meters per second."

Answer:

[tex]{P_2}=146.53\ kPa[/tex]

Explanation:

Volume ,V = 3 L

Initial temperature ,T₁ = 273 K

Initial pressure ,P₁ = 105 kPa

Final temperature ,T₂ = 381 K

Lets take final pressure =P₂

We know that ,If the volume of the gas is constant ,then we can say that

[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]

[tex]{P_2}=P_1\times \dfrac{T_2}{T_1}[/tex]

Now by putting the values in the above equation we get

[tex]{P_2}=105\times \dfrac{381}{273}\ kPa[/tex]

[tex]{P_2}=146.53\ kPa[/tex]

Therefore the final pressure will be 146.53 kPa.

The final pressure of the cylinder after being heated is 75.57 kPa.

To solve this problem, we can use the equation for Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature. In this case, the initial volume is 3.0 L and the initial temperature is 273 K. The final temperature is 381 K. Now we can set up a proportion:

(Initial volume) / (Initial temperature) = (Final volume) / (Final temperature)

Plugging in the numbers, we get:

(3.0 L) / (273 K) = (Final volume) / (381 K)

Solving for the final volume gives us:

Final volume = [(3.0 L) / (273 K)] x (381 K) =  4.1732 L

Since the volume remains the same, the pressure is inversely proportional to the new volume. So, if the initial pressure is 105 kPa, the final pressure can be calculated using the following equation:

(Initial pressure) x (Initial volume) = (Final pressure) x (Final volume)

Plugging in the numbers, we get:

(105 kPa) x (3.0 L) = (Final pressure) x (4.1732 L)

Solving for the final pressure gives us:

Final pressure = [(105 kPa) x (3.0 L)] / (4.1732 L) = 75.57 kPa

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Answer:

[tex]P_{abs}=107172.35Pa\\ P_{abs}=107.172kPa[/tex]

Explanation:

Given data

[tex]P_{a}=98kPa\\ p_{oil}=850kg/m^{3}\\ h=110cm=1.10m[/tex]

To find absolute pressure of air in the tank.We must find the pressure athe parallel point from the right tube

So

[tex]P_{abs}=P_{a}+hp_{oil}g\\P_{abs}=98000Pa+(1.10m*850kg/m^{3}*9.81 )\\P_{abs}=107172.35Pa\\ P_{abs}=107.172kPa[/tex]

Answer:

[tex]\vec{v}_{\rm max} = v\cos(\theta)(\^x)\\|\vec{v}_{\rm max}| = v\cos(\theta)\\\vec{a} = \vec{g} = -9.8\^y[/tex]

Explanation:

The equations of kinematics will be used to solve this question:

[tex]y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)\\v_y = v_{y_0} + a_yt[/tex]

At its maximum height, the projectile has zero velocity in the y-direction. But its velocity in the x-direction is unaffected.

First, let's apply the above equations to the x-direction.

There is no acceleration in the x-direction. So, its velocity in the x-direction is constant during the motion.

[tex]v_x = v_{x_0} + a_xt = v_{x_0} + 0\\v_x = v_{x_0} = v\cos(\theta)[/tex]

Therefore, the velocity vector of the projectile is

[tex]v_{max} = v_x = v\cos(\theta)[/tex]

The speed of the projectile is the same.

The acceleration vector is constant during the motion and equal to the gravitational acceleration, which is -9.8 downwards.

A Projectile motion fired upward at an angle with an initial speed has a velocity vector with horizontal and vertical components at its maximum height. Its speed at the maximum height is equal to the magnitude of the horizontal component of the velocity vector. The acceleration vector at the maximum height has a downward component due to gravity and no horizontal component.

When a projectile is fired upward at an angle $ heta$ above the horizontal with an initial speed $v_0$, at its maximum height, its velocity vector will have a horizontal component and a vertical component. The horizontal component of the velocity vector will remain constant throughout the motion, while the vertical component will be zero. The speed of the projectile at its maximum height will be equal to the magnitude of the horizontal component of its velocity vector. The acceleration vector at the maximum height will have a downward component due to gravity and no horizontal component.

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Answer:

933.804423995 m/s²

Explanation:

[tex]q_1[/tex] = Charge on particle 1 = 67 µC

[tex]q_2[/tex] = Charge on particle 2 = -25 µC

r = Distance between the particles = 47 cm

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

m = Mass of particle = 73 g

Electric force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times -25\times 10^{-6}\times 67\times 10^{-6}}{0.47^2}\\\Rightarrow F=-68.1677229516\ N[/tex]

The magnitude of force is 68.1677229516 N

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{68.1677229516}{0.073}\\\Rightarrow a=933.804423995\ m/s^2[/tex]

The acceleration is 933.804423995 m/s²

Answer:

1) True, 2) True, 3) False, 4) False, 5) False

Explanation:

1) True. Dissipative energy cannot be recovered, in general it is a form of heat

2) True. The dissipation can be by radiation, heat

3) False. Mechanical energy is divided into K and U but not in equal parts

4) False. When there are dissipative interactions, part of the mechanical energy is set in the form of heat, so its value decreases

5) False. Mechanical energy is the sum of those two energies

The correct statements are  a. The total mechanical energy of a system is constant only if nondissipative interactions occur.  and b.Mechanical energy can be dissipated to non-mechanical forms of energy. Mechanical energy is conserved only if no dissipative interactions occur, and it can be transformed into nonmechanical forms. Mechanical energy isn't always equally divided between kinetic and potential energy and isn't necessarily completely one or the other at any instant.

Let's analyze the provided statements about mechanical energy:

Thus, the correct statements are  a. The total mechanical energy of a system is constant only if nondissipative interactions occur.  and b.Mechanical energy can be dissipated to non-mechanical forms of energy.

Answer:

70 degree, 30 degree

Explanation:

We are given that [tex]\theta[/tex] is in standard position.

The reference angle [tex]\theta[/tex] is the acute angle.

Acute angle :The angle is always less than 90 degree.

The reference angle formed by terminal side of angle [tex]\theta[/tex] with x-axis.

When [tex]\theta=110^{\circ}[/tex]

Then, the reference angle =[tex]\bar{\theta}=180-110[/tex]

[tex]\bar{\theta}=70^{\circ}[/tex]

When [tex]\theta=210^{\circ}[/tex]

Therefore, the reference angle

[tex]\bar{\theta}=210-180[/tex]

[tex]\bar{\theta}=30^{\circ}[/tex]

In mathematics, a reference angle is the acute angle formed by the terminal side of a given angle and the x-axis, used for angles in standard position. The reference angle for θ = 110° is 70° and for θ = 210° it is 30°.

The concept of reference angles is often taught in mathematics, particularly in trigonometry to help understand the properties of angles in different quadrants of the Cartesian plane. A reference angle is always the acute angle (less than 90°) that the terminal side of the given angle makes with the x-axis. In standard position, the reference angle for an angle is the smallest angle between the terminal side of the angle and the horizontal axis (x-axis).

For an angle θ of 110°, which lies in the second quadrant, the reference angle would be 180° - 110° = 70°. Similarly, for an angle θ of 210° that lies in the third quadrant, the reference angle is calculated as 210° - 180° = 30°. Thus, the reference angle for θ = 110° is 70°, and for θ = 210°, it is 30°.