At the instant when the speed of the loop is 3.00 m/sm/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, [tex]q_1=6\ \mu C=6\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=2\ \mu C=2\times 10^{-6}\ C[/tex]

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

[tex]F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N[/tex]

So, the force applied between charges is 10.788 N.

Answer:

10.8 N

Explanation:

Answer:

B = 0.0404 т

Explanation:

Given

ΔE = 7.50 x 10⁻²⁵ J

μb = 927.4 x 10⁻²⁶ J/т

ΔE = 2 * μb * B

B = ΔE / 2 μb

B = 7.50 x 10⁻²⁵ /  2 * 927.4 x 10⁻²⁶

B = 7.50 / 1854.8 x 10⁻²⁵⁻⁽⁻²⁶⁾

B = 0.00404 x 10¹

B = 0.0404 т

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  [tex]\frac{460\times 32}{34\times 2}[/tex]

ω2  =  216.47

Final answer:

The distance between the reflected light beams can be found using reflection and refraction laws, applying Snell's law to calculate the angle of refraction, and subsequently using trigonometry to determine the light path within the glass and the shift caused by both reflections.

Explanation:

To determine the distance b between the light beam reflected from the top and the bottom surfaces of a glass plate, we need to use the laws of reflection and Snell's law for the refraction of light. The given information includes the thickness of the glass (2.8 cm), the index of refraction of the glass (1.6), and the angle of incidence (36°). When the laser beam strikes the top surface at an angle of incidence of 36°, it reflects at the same angle (law of reflection).

The angle of refraction can be found using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the air and glass, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively. Once the angle of refraction is determined, the actual path of the beam within the glass can be calculated using trigonometry, and subsequently the shift b caused by the refraction at the two interfaces (again using trigonometry).

Answer:

Explanation:

Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .

magnetic field in a torroid  B = μ₀ n I , n is number of turns per unit length and I is current .

B = 4π x 10⁻⁷ x (1000 / 2π x .16  )x 1 ( current = 1 A)

flux in the secondary coil

= B x area of face of coil x no of turns of secondary

= 4π x 10⁻⁷ x (1000 /2π x .16  ) .25 x 10⁻⁴ x 750

= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹

2343.75 x 10⁻⁸

= 23.43 x 0⁻⁶ H.

.

Answer:

2.5 x 10^-5 henry

Explanation:

The mutual inductance between the toroids is same.

mean radius of the toroid, r = 16 cm = 0.16 m

Area of crossection, A = 0.25 cm²

Number of turns in the first toroid, N1 = 1000

Number of turns in the second toroid, N2 = 750

The formula for the mutual inductance is given by

[tex]M =\frac{\mu_{0}N_{1}N_{2}A}{l}[/tex]

Where, l is the length

l = 2 x 3.14 x r = 2 x 3.14 x 0.16 = 1.0048 m

[tex]M =\frac{4\pi\times 10^{-7}\times 1000\times 750\times 0.25\times 10^{-4}}{1.0048}[/tex]

M = 2.5 x 10^-5 henry

Thus, the mutual inductance between the two toroid is 2.5 x 10^-5 henry.  

Answer:

Bouyant Force = 0.9N

The density of the jar and Penny was not given so we can't determine it's weigh and combined density

Explanation:

Bouyant force Fb = Volume of solid*density of liquid*acceleration

Fb= (200/2)*1*980.665

Fb= 98066.5/100000

Fb= 0.9N

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

Answer:

Magnitude of electric force = 0.03345 N

Explanation:

We are given;

Mass; m = 15g = 0.015kg

Angle above horizontal; θ = 35°

Speed; v = 6 m/s

Horizontal displacement; d = 2.9m

Now formula for time of flight is given as;

time of flight; t = (2Vsinθ)/g

Thus, plugging in values, we have

t = (2 x 6.0 x sin35)/9.8

t = (12 x 0.5736)/9.8

t = 0.7024 s

Now, let's find the acceleration

The formula for horizontal displacement is given by;

d = (Vcosθ)t + (1/2)at²

Plugging in the relevant values ;

2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²

2.9 = (4.2144 x 0.8192) + (0.2467)a

2.9 = 3.45 + (0.2467)a

(0.2467)a = 2.9 - 3.45

a = -0.55/0.2467

a = -2.23 m/s²

Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²

We know that F = ma

Thus,Force = 0.015kg x 2.23m/s² =

= 0.03345 N

Answer: [tex]1.8(10)^{-4} rad[/tex]

Explanation:

This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):

[tex]\theta=1.22\frac{\lambda}{D}[/tex]

Where:

[tex]\theta[/tex] is the angle of resolution (related to the acuity)

[tex]\lambda=554 nm=554(10)^{-9} m[/tex] is the wavelength of the light

[tex]D=3.75 mm=3.75(10)^{-3} m[/tex] is the diameter of the pupil

Solving:

[tex]\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}[/tex]

[tex]\theta=1.8(10)^{-4} rad[/tex]

The human eye's acuity is limited by light's diffraction through the pupil, which determines the minimum angle, Δθ, between two resolvable points. This is calculated using the formula Δθ = 1.22 (λ / D), where λ is the light's wavelength and D the diameter of the pupil.

The resolution of human visual acuity is related to the diffraction of light by the eye's pupil. This diffraction effect can be worked out using a formula that quantifies resolving power (Δθ), when the diameter of the aperture (in this case, the pupil size) and the wavelength of incident light are known.

In your example, a 3.75 mm diameter pupil encountering light of 554 nm wavelength can identify two points of light provided they subtend an angle (Δθ) at the pupil calculated by using the formula: Δθ = 1.22 (λ / D). For all light's range of visible wavelengths, the angle is extremely tiny - on the order of tens of thousands of a degree. This shows that the eye’s ability to distinguish two separate points of light is limited by the diffraction of light by the eye’s pupil.

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Answer:

(a) So range of frequency [tex]f > 43.27[/tex] Hz

(b) the reactance is 89.75 Ω

Explanation:

Given:

(a)

Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F

Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω

From the formula of reactance,

[tex]X_{C} = \frac{1}{\omega C}[/tex]

[tex]X_{C} = \frac{1}{2\pi fC}[/tex]

   [tex]f = \frac{1}{2\pi X_{C} C }[/tex]

   [tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]

   [tex]f = 43.27[/tex] Hz

So range of frequency [tex]f > 43.27[/tex] Hz

(b)

Capacitance [tex]C = 41 \times 10^{-6}[/tex] F

Frequency [tex]f = 43.27[/tex] Hz

From the formula of reactance,

[tex]X_{C} = \frac{1}{2\pi fC}[/tex]

[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]

[tex]X_{C} =[/tex] 89.75 Ω

Therefore, the reactance is 89.75 Ω

Answer: TRUE

Explanation:

Answer:

Explanation:

Velocity of sound wave at 30 degree = 350 m /s

frequency of sound = 300 Hz .

wavelength  of sound in air. = velocity / frequency

= 350 / 300

= 1.167 m

for wave formed in the rope :

velocity of wave in the rope

= [tex]\sqrt{\frac{T}{m} }[/tex]  

T is tension in the rope and m is mass per unit length .

m = .4 / 4

= .1

Putting the given values in the equation above

v = [tex]\sqrt{\frac{50}{.1} }[/tex]

v = 22.36 m /s .

velocity of wave in the rope.

= 22.36 m /s .

Answer:

The minimum wall thickness Tmin required for the spherical tank is 65.90mm

Explanation:

Solution

Recall that,

Tmin = The minimum wall thickness =PD/2бp

where D = diameter of 8.0 m

Internal pressure = 1.62 MPa

Then

The yield strength = 295MPa/3.0 = 98.33

thus,

PD/2бp = 1.62 * 8000/ 2 *98.33

= 12960/196.66 = 65.90

Therefore the wall thickness Tmin required for the spherical tank is 65.90mm

The data collected from all experiments by the students will be different. Option D is correct.

Kinetic energy:

The kinetic energy of the pendulum depends upon the length and mass and angle of the pendulum.

Since the angle is the same in all three experiments. Mass and length will decide the kinetic energy of the pendulum.

As students measure the kinetic energy against the function of time, the period of the pendulum will also vary.

The period of the pendulum can be calculated by the formula,

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

Where,  

[tex]l[/tex] - length  

[tex]g[/tex] - gravitational acceleration,

Therefore, the data collected from all experiments by the students will be different. Option D is correct.

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The motion of a simple pendulum is not affected by the mass of the bob; it is only influenced by the length of the pendulum and the acceleration due to gravity. If the pendulum lengths in the experiments are the same, then the kinetic energy data collected will be identical, even if the masses are different.

If three different experiments are conducted that pertain to the oscillatory motion of a pendulum with varying lengths and masses, and all are released from the same angle, the kinetic energy as a function of time for each experiment will differ only if the lengths of the pendulums are different. The mass of the pendulum bob does not affect the motion of a simple pendulum; pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity. Therefore, if the lengths of the pendulums in the different experiments are the same, then claim a. The data collected from Experiment 1 will be the same as the data collected from Experiment 2 is true irrespective of the different masses of the pendulum bobs.

The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity.

Answer:

[tex]9.38\times 10^7 m/s[/tex]

Explanation:

We are given that

Potential ,V=25 kV=[tex]25\times 10^3 V[/tex]

Distance,r =1 cm=[tex]\frac{1}{100}=0.01 m[/tex]

1 m=100 cm

Mass of electron, m=[tex]9.1\times 10^{-31} kg[/tex]

Charge, q=[tex]1.6\times 10^{-19} C[/tex]

We have to find the final velocity of the electron.

Speed of electron,[tex]v=\sqrt{\frac{2qV}{m}}[/tex]

Using the formula

[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}[/tex]

v=[tex]9.38\times 10^7 m/s[/tex]

Hence, the final velocity of the electron=[tex]9.38\times 10^7 m/s[/tex]

Answer:

see explanation

Explanation:

Net force along x axis is

[tex]\sum F_x = F \sin \theta= \frac{v^2}{R} ----(1)[/tex]

Net force along y axis is

[tex]\sum F_y = F \cos \theta= mg ----(2)[/tex]

The object can along accelerate down the stamp.

Thus F(net) is at the angle down the stamp

[tex]F_{net}=F_c\\\\F_{net}= \sin \theta mg\\\\F_c = \frac{mv^2}{r} \\[/tex]

where r = L in the direction of acceleration

[tex]\sin \theta mg = \frac{mv^2}{L}[/tex]

[tex]v^2 = gL \sin \theta[/tex]

[tex]v = \sqrt{gL \sin \theta}[/tex]

[tex]v = (gL \sin \theta )^{1/2}[/tex]

The relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].

The given parameters:

The relationship between the distance of the object, speed of the object can be determined by the net force on the toy is calculated follows;

[tex]Fsin(\theta) = \frac{mv^2}{L} \\\\mgsin(\theta) = \frac{mv^2}{L} \\\\gsin(\theta) = \frac{v^2}{L} \\\\v^2 = gL sin(\theta)\\\\v = \sqrt{gL sin(\theta)}[/tex]

Thus, the relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].

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Comets are divided into two types. Long-period comets are the comets that take more than two hundred years to finish an orbit throughout the Sun originate from the Oort Cloud.

Explanation:

Final answer:

A comet with a 324-year orbit is likely from the Oort Cloud, the distant spherical region of icy bodies surrounding our solar system, known as the source of long-period comets. The Kuiper Belt, on the other hand, sources short-period comets.

Explanation:

A comet with a period of 324 years is considered a long-period comet. These comets originate from a region far beyond the inner solar system. The Oort Cloud is a vast, spherical shell of icy bodies that extends up to about 50,000 astronomical units (AU) from the Sun and contains trillions of potential comets. This makes it the most likely source of long-period comets like the one mentioned. In contrast, the Kuiper Belt is a disk-shaped region beyond Neptune, extending to about 50 AU, which is known to source short-period comets, also known as Jupiter-family comets.

Long-period comets like the one with a 324-year orbit are thought to be influenced by gravitational perturbations from nearby passing stars or galactic tides, which can send them towards the inner solar system. Once these comets enter the inner solar system, they can have dramatic interactions with planets or the Sun, sometimes leading to their disintegration or alteration in orbit. An example of such an interaction was when Comet Shoemaker-Levy 9 broke apart and collided with Jupiter in 1994.

Answer:

Answer

Explanation:

In a LCR Circuit, the phase difference between voltage and current is usually summarized as;

tan∅ = XL- XC

               R

     =WL - 1/WC

            R

When w=0 ( i.e when it is very low)    tan∅ = ∞

                                                         or      ∅ = -90

When w=0 ( i.e when it is very large)    tan∅ =+ ∞

                                                         or         ∅ = +90

Answer:

1.04x[tex]10^{-3}[/tex] s

Explanation:

->The maximum current through the resistor is

[tex]I_{max}[/tex] = V/R = V/[tex]Re^{-t/RC}[/tex]= V/R×[tex]e^{0}[/tex] = V/R

Voltage 'V'=50V

Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω

Therefore, [tex]I_{max}[/tex]=50/35=> 1.43 A

->The maximum charge can be determined by

Q = CV

where,

Capacitance of the capacitor 'C' = 30.0µF = 30×10-⁶F

Therefore,

Q=30×10-⁶ x 50=>1.5 x [tex]10^{-3}[/tex]

In order to find that when does the maximum current occur, the time taken given the quantity of charge and the electric current is:

t= Q / I=>  1.5 x [tex]10^{-3}[/tex]/ 1.43

t= 1.04x[tex]10^{-3}[/tex] s

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

The spring has been extended for 3.5 m

Explanation:

We have the formula,

PE =1/2 K X²

Rewrite the equation as

PE=1/2 K d²

multiply both the sides by 2/K to simplify the equation

2/k . PE= 1/2 K  d² . 2/K

√d²=√2PE/K

Cancelling the root value and now we have,

d=√2PE/k

d=√2×98 J / 16N/m

d=√12.25

d=3.5 m

The spring has been extended for 3.5 m

Final answer:

To find the extension of a spring with a constant of 16 N/m and 98 J of energy stored, we use the formula for potential energy. The spring is extended by 3.5 meters.

Explanation:

The question involves calculating the extension of a spring based on the energy stored in it and the spring's constant. The formula for the potential energy stored in a spring is given by

U = 1/2 kx², where U is the potential energy, k is the spring constant, and x is the extension of the spring from its equilibrium position. In this case, the energy (U) is 98 J and the spring constant (k) is 16 N/m.

The spring in question has an energy of 98 J stored in it.

Using the formula for potential energy stored in a spring, we have:

[tex]PE = 1/2 k x^2[/tex]

By substituting the known values, we find that the spring is extended by 0.7 m (70 cm

We rearrange the formula to solve for x:

x = √(2U/k).

Substituting the given values,

x = √(2*98/16)

= √(196/16)

= √(12.25)

= 3.5 meters. Therefore, the spring is extended by 3.5 meters.

Answer:

This problem is incomplete, it says like this: The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You dont find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 N/m that will compress with a force of 4400 N. The inside of the gun barrel is coated with teflon, so the average friction force will be only 40 N during the 4 m his moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 m above his initial rest position?

The speed is 15.5 m/s, and the image shows the vector drawing.

Explanation:

For the calculation of the force that is due to the spring is equal to:

F = kx

Where

F = 4400 N

k = 1100 N/m

x = F/k = 4400/1100 = 4 m

For the calculation of the total energy is equal to:

[tex]\frac{1}{2} kx^{2} =\frac{1}{2}mv^{2} +mgh+Fx\\v=\sqrt{\frac{2}{m} (\frac{1}{2}kx^{2} -mgh-Fx) }[/tex]

Where

m = 60 kg

h = 2.5 m

Replacing:

[tex]v=\sqrt{\frac{2}{60}(\frac{1}{2}*1100*4^{2}-60*9.8*2.5-(40*4)) } =15.5m/s[/tex]

The problem is about setting coordinate axes and vectors for a physics problem, with the origin placed at the performer's initial position. The vectors represent various physical quantities depending on their orientation and direction, always starting from the origin.

This is a typical problem in physics dealing with vectors, coordinate axes, and natural phenomena like gravitational potential energy. Setting up the appropriate coordinate system is crucial. We set the origin, or the (0,0) point, at the performer's initial position as mentioned in the problem.

The y-axis should be vertical, aligned with the direction of gravity. This makes computations involving gravity much easier. It's important to note that increasing height (direction opposite to that of gravity) is typically represented as positive y.

The vectors originating from the black dot (our origin) represent quantities such as displacement, velocity, force etc. The direction of the vector indicates the direction of the quantity.

For example, if there's a vector pointing directly upwards from the origin, it could represent an upward force or movement. Keep in mind, the orientation and location of the vectors are determined by the physical reality the problem is describing, and they should always start from the origin in the problem's context.

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The magnitude of the induced electric field can be found using Faraday's law of electromagnetic induction. The current flows clockwise as viewed by someone on the south pole of the magnet.

To find the magnitude of the electric field induced in the ring, we can use Faraday's law of electromagnetic induction. According to the law, the magnitude of the induced electric field is equal to the rate of change of magnetic flux through the ring. The magnetic flux can be calculated by multiplying the magnetic field strength by the area of the ring. In this case, the magnetic field is decreasing at a rate of 0.260 T/s, and the area of the ring is π(2.50 cm)2. Therefore, the magnitude of the induced electric field is 2π(2.50 cm)2(0.260 T/s).

As for the direction of the current flow, it can be determined using the right-hand rule for induced currents. If you point your right thumb in the direction of the magnetic field (from north to south), the induced current will flow in the direction of your curled fingers, which in this case will be clockwise as viewed by someone on the south pole of the magnet.

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θ₃ is 27.12°

Explanation:

Using Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

where,

n₁ = refractive index of material with incident light

n₂ = refractive index of material with refracted light

(a)

1 → air

2 → glas

3 → water

n(water) = n₃ = 1.333

From air to glass,

n₁ sin θ₁ = n₂sin θ₂

The angle of refraction 2 is the angle of incidence when the light comes from glass to water.

From glass to water:

n₁ sin θ₁ = n₃ sin θ₃

1 X sin 37.5° = 1.333 sin θ₃

sin θ₃ =  sin (37.5)/ 1.333

sin θ₃ = 0.456

θ₃ = 27.12°

Therefore, θ₃ is 27.12°

Answer:

a)  Vki = 0.4 J  

b) v = 0.98 m/s

c)  final position = 0.11745 m from initial position

Explanation:

Given:-

- The mass of block, m = 500 g

- The spring constant, k = 80 N/m

- The coefficient of the surface, u = 0.20

- The inclination of the block, θ = 30°

- The block is initially compressed, xi = 10 cm

- The block is initially at rest, vi = 0 m/s

Find:-

(a) How much potential energy was stored in the block-spring-support system when the block was just released?

(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.

(c) Determine the position of the block where it just comes to rest on its way up the incline.

Solution:-

- The potential energy initially stored in a spring "Vki" with constant "k" which has undergone a displacement of "x" is given by:

                            Vki = 0.5*k*xi^2

                            Vki = 0.5*80*0.1^2

                            Vki = 0.4 J  

- The block is released from rest when the energy stored in the spring is dispensed in three forms of energy.

- There would be an increase in the potential energy " Ep " of the block as it moves up.

- There would be an increase in kinetic energy for some time " Ek "

- There would be loss of total energy due to fictitious force ( Ff) the work done against friction will dissipate energy.

- The increase in potential energy "Ep" at displacement xo = 0 from mean position is given by:

                         ΔEp = m*g*Δh ... change in vertical height h

                         ΔEp = m*g*( xi - xo )*sin ( θ )

                         ΔEp = 0.5*9.81*(0.1)*sin ( 30 )

                         ΔEp = 0.24525 J

- The increase in kinetic energy "Ek" at displacement xi = 10 cm, vi = 0 m/s from initial position to mean position at xo = 0 ,its velocity is "vf" is given by:

                         ΔEk = 0.5*m*( vf^2 - vi^2 ) ... change in velocity

                        ΔEk = 0.5*m*( vf )^2

- There would be loss of total energy due to fictitious force ( Ff ) the work done against friction will dissipate energy. First apply the equilibrium conditions on the block normal to slope and determine the contact force ( Nc ):

                        Nc - m*g*cos ( θ ) = 0

                        Nc = m*g*cos ( θ )

- The friction force ( Ff ) is given by:

                        Ff = Nc*u

                        Ff = u*m*g*cos ( θ )

- The work done by block against friction is given by:

                        Wf = -Ff*( xi - xo )

                        Wf = -u*m*g*xi*cos ( θ )

                        Wf = -0.2*0.5*9.81*0.1*cos ( 30 )  

                        Wf = -0.08495 J

- We can now express the work done principle for the block:

                        Vki = ΔEp + ΔEk + Wf

                        ΔEk = - ΔEp - Wf + Vki

                        0.5*m*( vf )^2 = -0.24525 + 0.08495 +0.4

                        vf^2 = 4*(-0.24525 + 0.08495 +0.4 )

                        vf = √0.9588

                        vf = 0.98 m/s

- We will denote the extension of spring at top most position from mean position as "x".

- From mean position xo = 0 m. The block will further move up the slope and expense all its kinetic energy "Ek" in the form of gain in potential energy and gain in elastic potential energy "Vk" and work is done against friction.

               Vki = Ep2 + Wf + Vk2

              0.4 = mg*x*sin ( 30 ) + 0.24525 + 0.08495 + u*m*g*x*cos ( θ ) + 0.5*k*x^2

               0.5*80*x^2 + x*(0.5*9.81*sin(30) + 0.2*0.5*9.81*cos ( 30 ) ) + 0.3302-0.4 =0

               40x^2 + 3.30207x - 0.0698 = 0

Solve the quadratic equation:

               x = -0.1 m (10 cm) - initial compression at rest ;

               x = 0.01745 m

- So the extension in spring at the rest position is x = 0.01745 m. The position of the next resting point is:

                final position = xi + x

                final position = 0.1 + 0.01745

                final position = 0.11745 m from initial position.

                x =  

(a) The stored PE will be "0.40 J".

(b) The block's speed will be "0.48 m/s".

(c) The block's position will be at "12 cm".

According to the question,

Mass, m = 500 h

Spring constant = 80 N/m

Coefficient of friction = 0.20

Inclined angle = 30°

(a) We know the formula,

Spring P.E = [tex]\frac{1}{2}[/tex] kx²

By substituting the values,

                 = [tex]\frac{1}{2}[/tex] × 80 × (0.1)²

                 = 0.40 J

(b) We know that,

→ [tex]P.E_s[/tex] = ΔKE + Work done by friction + Δ[tex]PE_g[/tex]

or,

          = [tex]\frac{1}{2}[/tex] mv² + [tex]\mu_R[/tex] mgx Cosθ + mgx Sinθ

By substituting the values,

 0.40 = [tex]\frac{1}{2}[/tex] × 0.5 × v² + 0.2 × 0.5 × 9.8 × Cos30° × 0.1 + 0.5 × 9.8 × 0.1 ×  Sin30°

          = 0.48 m/s

(c) By using the above formula,

→ 0.40 = [tex]\mu_R[/tex] mgl Cosθ + mgl Sinθ

           = 0.2 × 0.5 × 9.8 × Cos30° × l + 0.5 × 9.8 × l × Sin30°

         l = 0.12 m or,

           = 12 cm

Thus the above responses are correct.

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Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

Answer:

first ball is at rest after collision and the kinetic energy of the first ball is transferred to the second ball after collision.

Explanation:

mass of each ball = m

initial velocity of first ball = u

initial velocity of second ball = 0 m/s

final velocity of second ball = initial velocity of first ball = u

Use conservation of momentum

let v is the final velocity of the first ball after collision

m x u + m x 0 = m x v + m x u

So, v = 0

It means after the collision the second ball moves with the velocity which is equal to the initial velocity of the first ball and the first ball comes at rest.

As there is no friction between the balls during the collision, the collision of two balls is perfectly elastic and thus the kinetic energy of the system is conserved.

The entire kinetic energy of the first ball is transferred to the second ball after the collision.

The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.

According to the law of conservation of momentum, the change in momentum of a body before the collision is equal to the change in momentum after the collision.

The formula for calculating momentum is expressed as:

[tex]\rho = mv \\[/tex]

According to the conservation of momentum

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

Since ball 2 is initially at rest [tex]u_2=0[/tex]

Also if after the collision, ball 2 departs with the same velocity that ball 1 originally had, hence [tex]v=u_1[/tex]

Substituting the given parameters into the formula:

[tex]m_1u_1+m_2(0)=(m_1+m_2)u_1\\m_1u_1=m_1u_1+m_2u_2\\m_2u_2=0[/tex]

This shows that the velocity of the second ball is also zero

Due to the absence of friction between the colliding object, the collision is elastic in nature showing that the kinetic energy of the system is conserved.

The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.

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Answer:

the hospital needs to order for a  minimum amount of 0.4 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex]

Explanation:

Given that:

A hospital needs 0.100 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex]

and it takes 10 days for the shipment to arrive:

the half life [tex]t_{1/2}[/tex] = 5 days

So, since the half life = 5 days ;

decay constant [tex]\lambda = \frac{In_2}{t_{1/2}}[/tex]

where:

[tex]N_o= ??? \\ \\ N = 0.1 00 \\ \\ t = 10 days \\ \\ N(t)= N_oe^{-\lambda t}\\ \\0.1 = N_oe^{\frac{-In_2}{5} *10} \\ \\0.1 = N_oe^{ - In \ 4} \\ \\ 0.1 = N_oe^{ In \frac{1}{ 4}} \\ \\ 0.1 = \frac{N_o}{4} \\ \\ N_o = 0.1*4 \\ \\ N_o = 0.4 \ g[/tex]

Therefore in order to get 0.100 g  of [tex]\left \ {{133} \atop {54} \right. Xe[/tex], the hospital needs to order for a  minimum amount of 0.4 g of

Answer:

Explanation:

1. They have different wavelengths - Because These radiations form a spectra that differs by the size of the wavelength.

2. They have different Frequencies (f) that is frequency = 1/ wavelength

(f = 1/wavelength)

3. They propagate at different speed though a non vacuum media (non vacuum media affect the speed based on the wavelength)

Answer:

a) [tex]E_{tot} = 7\,J[/tex], b) [tex]x = \pm\sqrt{\frac{7}{4} }[/tex]c) [tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]

Explanation:

a) The total energy of the object is:

[tex]E_{tot} = U + K[/tex]

[tex]E_{tot} = 4\cdot x^{2} + \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]E_{tot} = 4\cdot (-0.50\,m)^{2} + \frac{1}{2}\cdot (3\,kg)\cdot (2\,\frac{m}{s} )^{2}[/tex]

[tex]E_{tot} = 7\,J[/tex]

b) The total energy of the object is:

[tex]E_{tot} = U[/tex]

[tex]7\,J = 4\cdot x^{2}[/tex]

[tex]x = \pm\sqrt{\frac{7}{4} }[/tex]

c) The speed of the object is clear in the total energy expression:

[tex]E_{total} = U + K[/tex]

[tex]K = E_{total}-U[/tex]

[tex]\frac{1}{2}\cdot m \cdot v^{2} = E_{total} - 4\cdot x^{2}[/tex]

[tex]v^{2} = \frac{2\cdot (E_{total}-4\cdot x^{2})}{m}[/tex]

[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]

[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0.6\,m)^{2}]}{3\,kg} }[/tex]

[tex]v \approx 1.925\,\frac{m}{s}[/tex]

The magnitude of the momentum is:

[tex]p = (3\,kg)\cdot (1.925\,\frac{m}{s} )[/tex]

[tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]

d) Before calculating the acceleration experimented by the object, it is required to determine the net force exerted on it. There is a relationship between potential energy and net force:

[tex]F = -\frac{dU}{dx}[/tex]

[tex]F = -8\cdot x[/tex]

Acceleration experimented by the object is:

[tex]a = -\frac{8\cdot x}{m}[/tex]

[tex]a = -\frac{8\cdot (0.6\,m)}{3\,kg}[/tex]

[tex]a = -1.6\,\frac{m}{s^{2}}[/tex]

e) The position of the object versus time is found by solving the following differential equation:

[tex]\frac{d^{2}x}{dt} +\frac{8\cdot x}{m} = 0[/tex]

[tex]s^{2}\cdot X(s)- s\cdot v(0) - x(0) + \frac{8}{m}\cdot X(s) = 0[/tex]

[tex](s^{2} + \frac{8}{m})\cdot X(s) = s\cdot v(0)+x(0)[/tex]

[tex]X(s) = \frac{s\cdot v(0)+x(0)}{(s^{2}+\frac{8}{m} )}[/tex]

[tex]X(s) = v(0)\cdot \frac{s}{s^{2}+\frac{8}{m} } +\frac{m\cdot x(0)}{8} \cdot \frac{\frac{8}{m}}{s^{2}+\frac{8}{m}}[/tex]

[tex]x(t) = v(0) \cdot \cos \left(\frac{8}{m}\cdot t \right)+\frac{m\cdot x(0)}{8}\cdot \sin \left(\frac{8}{m}\cdot t \right)[/tex]

The velocity is obtained by deriving the previous expression:

[tex]v(t) = -\frac{8\cdot v(0)}{m}\cdot \sin \left(\frac{8}{m}\cdot t \right)+x(0)\cdot \cos \left(\frac{8}{m}\cdot t \right)[/tex]

Speed of the object at [tex]x = 0[/tex] is:

[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]

[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0\,m)^{2}]}{3\,kg} }[/tex]

[tex]v \approx 2.160\,\frac{m}{s}[/tex]

The equation of the motion are:

[tex]x(t) = 2.160\cdot \cos \left(2.667\cdot t \right)[/tex]

[tex]v(t) = -5.76\cdot \sin (2.667\cdot t)[/tex]

The expression for the kinetic energy of the object is:

[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]K = \frac{1}{2}\cdot (3\,kg)\cdot [33.178\cdot \sin^{2}(2.667\cdot t)][/tex]

[tex]K = 49.767\cdot \sin^{2}(2.667\cdot t)[/tex]

Graphics are included below as attachments. (Position versus time, kinetic energy vs time).

Following are the responses to the given points:

Given:

object mass [tex](m)= 3.0\ kg \\\\[/tex]

Potential energy [tex]U(x)=4.0\ x^2\\\\[/tex]  

[tex]\to x=-0.5 \ m\\\\ \to velocity \ (v)=2.0 \ \frac{m}{s}\\\\[/tex]

Solution:

For point (a)  

Total Energy [tex](TE) = PE + KE \\\\[/tex]

[tex]\to PE = 4.0 (0.5 m)^2 = 1\ J\\\\ \to KE = \frac{1}{2} mv^2 = 0.5 \times 3.0\ kg (2.0 \frac{m}{s})^2 = 6\ J \\\\\to TE = 7\ J\\\\[/tex]

For point (b)  

If an object's potential energy is 7 J, it has 0 kinetic energy.

[tex]\to U(x) = 4x^2 = 7\ J \\\\\to x=+1.2 \ m, -1.2 \ m[/tex]

For point (c)  

[tex]\to x=0.60\ m \\\\ \to TE=4x^2 + \frac{1}{2}mv^2 = 7\ J\\\\ \to 4(0.60)^2 +0.5 \times 3.0 \ kg \times v^2 = 7\ J \\\\ \to v=1.92 \ \frac{m}{s}\\\\ \to Momentum (p) = mv \\\\\to p= 3.0\ kg \times 1.92 \ \frac{m}{s}\\\\ \to p= 5.76 kg \ \frac{m}{s}\\\\[/tex]

For point (d)

[tex]\to x_1= 0.6 \ m\\\\ \to v_1 = 1.92 \frac{m}{s}\\\\ \to x_2 = 1.2\ m[/tex] the velocity is found by  

[tex]\to 4x^2 + \frac{1}{2} mv^2= 7\ J \\\\ \to 4(1.2)^2 +0.5 \times 3.0\ kg \times v^2 = 7\ J \\\\ \to v_2 = 0.9 \ \frac{m}{s}\\\\[/tex]

Calculating the acceleration:

[tex]\to V^2_{2}-v^2_{1}= 2a(x_2-x_1) \\\\\to (0.9 \ \frac{m}{s})^2 - (1.92\ \frac{m}{s})^2 = 2 a(1.2\ m - 0.6\ m) \\\\\to a= -2.4 \frac{m}{s^2}\\\\[/tex]

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