A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{1}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{1}=0.01021\ Wb[/tex]

We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{2}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{2}=0.00127\ Wb[/tex]

We need to calculate the magnitude of the average induced emf

Using formula of emf

[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]

Put the value into t5he formula

[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]

[tex]\epsilon=0.016867\ V[/tex]

[tex]\epsilon=16.87\ mV[/tex]

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

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

Put the value into the formula

[tex]I=\dfrac{0.016867}{2.5}[/tex]

[tex]I=0.00675\ A[/tex]

[tex]I=6.75\ mA[/tex]

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

Answer: (V1+V2)/2

Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2

Answer:

(a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .

Explanation:

Given that,

Width of diffraction grating [tex]w= 19.2\ mm[/tex]

Number of rulings[tex]N=6010[/tex]

Wavelength = 337 nm

We need to calculate the distance between adjacent rulings

Using formula of distance

[tex]d=\dfrac{w}{N}[/tex]

Put the value into the formula

[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]

[tex]d=3.19\times10^{-6}\ m[/tex]

We need to calculate the value of m

Using formula of constructive interference

[tex]d \sin\theta=m\lambda[/tex]

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

Here, m = 0,1,2,3,4......

[tex]\lambda[/tex]=wavelength

For largest value of  θ

[tex]\dfrac{m\lambda}{d}>1[/tex]

[tex]m>\dfrac{d}{\lambda}[/tex]

Put the value into the formula

[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]

[tex]m>9.46[/tex]

[tex]m = 9[/tex]

(a). We need to calculate the largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=71.9^{\circ}[/tex]

(b). We need to calculate the second largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=57.7^{\circ}[/tex]

(c). We need to calculate the third largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=47.7^{\circ}[/tex]

Hence, (a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .

Answer:

Not sure but I believe predictable.

Explanation:

Phenomena usually consists of :

- A history, a date in which the physical phenomenon has occurred.

- A source, a place or reason to why or where the physical phenomena has occured.

According to this, I want to say predictable.

It is not repeatable, there are one-time phenomenons that have occurred that scientists to this day still have not recorded again such as the Big Bang.

It is not provable. Most of the theories earlier scientists and historians have predicted have not yet been proved. Yet rather, somehow, they have been explored and investigated.

It is not readable. This is self explanatory, some things scientists investigate are not written down, nor read. It starts with a mental theory and then immediately goes to the next phase of investigation.

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

Answer:

The acceleration is   [tex]a =2\ m/s^2[/tex]

Explanation:

From the question we are told that

       The  mass of the jumbo jet is  [tex]m_j = 100000\ kg[/tex]

        The thrust is  [tex]F_k = 50000 \ N[/tex]

Generally given that the jet has four engines the total thrust is  

        [tex]F_t = 4 * F_k[/tex]

substituting values

       [tex]F_t = 4 * 50000[/tex]

      [tex]F_t = 200000 \ N[/tex]

Generally the acceleration of the is mathematically represented as

         [tex]a = \frac{F_t}{m}[/tex]

substituting values

       [tex]a =2 \frac{N}{kg}[/tex]

Now  

        [tex]N = kg \cdot m/s^2[/tex]

Hence

         [tex]a =2 \frac{kg * \cdot m/s^2}{kg}[/tex]

        [tex]a =2\ m/s^2[/tex]

Answer:

[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.

Explanation:

According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:

[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]

[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]

[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.

Answer:

The number of interference fringes is  [tex]n = 3[/tex]

Explanation:

From the question we are told that

     The wavelength is  [tex]\lambda = 433 \ nm = 433 *10^{-9} \ m[/tex]

      The distance of separation is  [tex]d = 6 \mu m = 6 *10^{-6} \ m[/tex]

       The  order of maxima is m =  5

The  condition for constructive interference is

       [tex]d sin \theta = n \lambda[/tex]

=>     [tex]\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ][/tex]

=>    [tex]\theta = 21.16^o[/tex]

So at  

      [tex]\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m[/tex]

   [tex]6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}[/tex]

=>    [tex]n = 3[/tex]

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

Answer:

[tex]N = 208 \ turns[/tex]

Explanation:

From the question we are told that

    The  rate of  current change is  [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]

    The  magnitude of the induced emf is  [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]

     The  current is  [tex]I = 1.50 \ A[/tex]

      The  average  flux is  [tex]\phi = 0.00458 \ Wb[/tex]

Generally the number of  turns the number of turn the solenoid has is mathematically represented as  

            [tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]

substituting values

           [tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]

            [tex]N = 208 \ turns[/tex]

Answer:

8 possible combinations

Assuming R 1, R 2 and R 3 be three different Resistance

1- all three in series

2-all three in parallel

3- R 1 and R 2 in series and parallel with R 3

​

4-R 1 and R 3 in series and parallel with R 2

​

5-R 2 and R 3 in series and parallel with R 1

​

6- R 1 and R 2

in parallel and series with R 3

​

7-R 1 and R 3 in parallel and series with R 2

​

8-R 2 and R 3 in V with R 1

​

Answer:

196 nm

Explanation:

Given that

Value of wavelength, = 490 nm

Time spent in air, t(a) = 17.5 ns

Thickness of glass, th = 0.8 m

Time spent in glass, t(g) = 21.5 ns

Speed of light in a vacuum, c = 3*10^8 m/s

To start with, we find the difference between the two time spent

Time spent on glass - Time spent in air

21.5 - 17.5 = 4 ns

0.8/(c/n) - 0.8/c = 4 ns

Note, light travels with c/n speed in media that has index of refraction

(n - 1) * 0.8/c = 4 ns

n - 1 = (4 ns * c) / 0.8

n - 1 = (4*10^-9 * 3*10^8) / 0.8

n - 1 = 1.2/0.8

n - 1 = 1.5

n = 1.5 + 1

n = 2.5

As a result, the wavelength of light in a medium with index of refraction would then be

490 / 2.5 = 196 nm

Therefore, our answer is 196 nm

Answer: 156.02 metre.

Explanation:

Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.

Let us use third equation of motion,

V^2 = U^2 + 2as

Since the car is decelerating, V = 0

And acceleration a will be negative.

U = 33 mph

S = 39 m

Substitute both into the formula

0 = 33^2 - 2 × a × 39

0 = 1089 - 78a

78a = 1089

a = 1089 / 78

a = 13.96 m/h^2

If we assume that the car decelerate at the same rate.

the distance the car will stop had it been going 66.0mph will be achieved by using the same formula

V^2 = U^2 + 2as

0 = 66^2 - 2 × 13.96 × S

4356 = 27.92S

S = 4356 / 27.92

S = 156.02 m

Therefore, the car would stop at

156.02 m

Answer:

The current is  [tex]I = 1420 \ A[/tex]

Explanation:

From the question we are told that

   The  diameter of the wire is  [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]

    The  magnetic field is  [tex]B = 0.100 \ T[/tex]

Generally the radius of the wire is mathematically evaluated as

       [tex]r = \frac{d}{2}[/tex]

substituting values

     [tex]r = \frac{ 0.00568}{2}[/tex]

     [tex]r = 0.00284 \ m[/tex]

Generally the magnetic field is mathematically represented as

       [tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]

=>    [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space  with value [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]

substituting values

=>     [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]

=>     [tex]I = 1420 \ A[/tex]

Complete Question

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .

Part A. Calculate  the coil's self-inductance.

Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.

Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?

Answer:

Part A  

       [tex]L = 0.000863 \ H[/tex]

Part B  

       [tex]\epsilon = 0.863 \ V[/tex]

Part C

    From terminal a to terminal b

Explanation:

From the question we are told that

      The  number of turns is  [tex]N = 590 \ turns[/tex]

      The cross-sectional area is  [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]

      The  radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]

Generally the coils self -inductance is mathematically represented as

              [tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]

Where [tex]\mu_o[/tex] is the permeability of  free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting values

             [tex]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]

             [tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]

             [tex]L = 0.000863 \ H[/tex]

Considering the Part B

      Initial current is [tex]I_1 = 5.00 \ A[/tex]

      Current at time t is [tex]I_t = 3.0 \ A[/tex]

       The  time taken is  [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]

The self-induced emf is mathematically evaluated as

          [tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]          

=>         [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]

substituting values

             [tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]  

             [tex]\epsilon = 0.863 \ V[/tex]

The direction of the induced emf is  from a to b because according to Lenz's law the induced emf moves in the same direction as the current

This question involves the concepts of the self-inductance, induced emf, and Lenz's Law

A. The coil's self-inductance is "0.863 mH".

B. The self-induced emf in the coil is "0.58 volts".

C. The direction of the induced emf is "from b to a".

A.

The self-inductance of the coil is given by the following formula:

[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]

where,

L = self-inductance = ?

[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²

N = No. of turns = 590

A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²

r = radius = 5 cm = 0.05 m

Therefore,

[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]

L = 0.863 x 10⁻³ H = 0.863 mH

B.

The self-induced emf is given by the following formula:

[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]

where,

E = self-induced emf = ?

ΔI = change in current = 2 A

Δt = change in time = 3 ms = 0.003 s

Therefore,

[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]

E = 0.58 volts

C.

According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.

Learn more about Lenz's Law here:

https://brainly.com/question/12876458?referrer=searchResults

Answer:

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

Answer: no u

Explanation: no u

Answer:

D. evacuated tube collector due to reduced heat loss

Explanation:

Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.

During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.

Answer:

y = 20.99 V / A

there is no friction    y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight  hydrostatic thrust

              F_net= ∑F = B-W

the expression stop pushing is

              B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

             W = ρ-ice g V

              F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

            F = -dU / dy

            ΔU = - ∫ F dy

            ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

            ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0,  U = 0, that is, the potential energy is zero at the surface

             U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

             Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

             [tex]Em_{f}[/tex] = U = m g y

as there is no friction, energy is conserved

            Em₀ = Em_{f}

            (ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

            ρ_ice = m / V

            m = ρ_ice V

we substitute

             (ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

              y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

             y = 0.913 / (1 - 0.913) 2 V / A

             y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

                V / A = h

where h is the height of the block

                 y = 20.99 h

Answer:

The wavelength is  [tex]\lambda = 419 \ nm[/tex]

Explanation:

From the question we are told that

   The  distance of separation is   [tex]d = 3.34 *10^{-5} \ m[/tex]

   The  distance of the screen is  [tex]D = 3.30 \ m[/tex]

      The  order of the fringe is  n =  7

     The distance of separation of  fringes is y =  29.0 cm = 0.29 m

Generally the wavelength of the light is mathematically represented as

          [tex]\lambda = \frac{y * d }{ n * D}[/tex]

substituting values

         [tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]

        [tex]\lambda = 4.19*10^{-7}\ m[/tex]

        [tex]\lambda = 419 \ nm[/tex]

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

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

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

Answer:

The question is not clear enough. So i have attached a copy of the correct question.

A) B = V/c

B) c = Lv

Explanation:

A) we know that formula for magnetic flux is;

Φ = BA

Where B is magnetic field and A is area

Now,

Let's differentiate with B being a constant;

dΦ/dt = B•dA/dt

From faradays law, the EMF induced is given as;

E = -dΦ/dt

However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.

Thus, V = -dΦ/dt

Thus, we can now say that;

-V = B•dA/dt

Now from the question, we are told that dA/dt = - c

Thus;

-V = B•-c

So, V = Bc

Thus, B = V/c

B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:

Thus;

E =LvB or can be written as; V = LvB

Where;

V is EMF

L is length of bar

v is velocity

From the first solution, we saw that;

V = Bc

Thus, equating both of the equations, we have;

Bc = LvB

B will cancel out to give;

c = Lv

Explanation:

Answer:

Using a process of testing ideas and gathering evidence

Explanation:

Answer:

a)   Δt = 24.96 s , b)  τ = 0.078 N m

Explanation:

This is a rotational kinematics exercise

        θ = w₀ t - ½ α t²

Let's reduce the magnitudes the SI system

       θ = 60 rev (2π rad / 1 rev) = 376.99 rad

       w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s

      α = (w₀ t - θ) 2 / t²

let's calculate the annular acceleration

      α = (43.98 10 - 376.99) 2/10²

      α = 1,258 rad / s²

Let's find the time it takes to reach zero angular velocity (w = 0)

        w = w₀ - alf t

         t = (w₀ - 0) / α

         t = 43.98 / 1.258

         t = 34.96 s

this is the total time, the time remaining is

         Δt = t-10

         Δt = 24.96 s

To find the braking torque, we use Newton's law for angular motion

        τ = I α

the moment of inertia of a circular ring is

       I = M r²

we substitute

         τ = M r² α

we calculate

        τ = 0.625  0.315²  1.258

        τ = 0.078 N m

The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Given data:

The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex]   (rps means rotation per second).

The time interval is, t' = 10.0 s.

The number of rotations made by wheel is, n = 60.0.

The mass of bike wheel is, m = 0.625 kg.

The radius of wheel is, r = 0.315 m.

The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,

[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]

Here, [tex]\theta[/tex] is the angular displacement, and its value is,

[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]

And, angular speed is,

[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]

Solving as,

[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]

Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.

[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]

Then total time is,

T = t - t'

T = 35.18 - 10

T = 25.18 s

Now, use the standard formula to obtain the value of braking torque as,

[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]

Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Learn more about the rotational motion here:

https://brainly.com/question/1388042

Explanation:

Hey, there!

Here, In question given that,

potential difference (V)= 8V

resistance (R)= 10 ohm

Now,

According to the Ohm's law,

V= R×I { where I = current}

or, I = V/R

or, I = 8/10

Therefore, current is 4/5 A or 0.8 A.

(A= ampere = unit of current).

Hope it helps...

Answer:

The  rotational kinetic energy is  [tex]K = 2116.3 \ J[/tex]

Explanation:

From the question we are told that

    The moment of inertia  is  [tex]I = 0.35 \ kg \cdot m^2[/tex]

    The number of revolution is N  =  70 revolution

     The  time taken is  t  =  4.0  s

Generally the angular velocity is mathematically represented as

      [tex]w = \frac{2 \pi N }{t }[/tex]

substituting values

      [tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]

       [tex]w = 109.97 \ rad/s[/tex]

The rotational kinetic energy K i mathematically represented as

       [tex]K = \frac{1}{ 2} * I * w^2[/tex]

substituting values

       [tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]

       [tex]K = 2116.3 \ J[/tex]