Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.

Answer:

The magnetic field is [tex]B = 8.20 *10^{-3} \ T[/tex]

Explanation:

From the question we are told that

   The  mass of the metal rod is  [tex]m = 0.12 \ kg[/tex]

    The current on the rod is  [tex]I = 4.1 \ A[/tex]

    The distance of separation(equivalent to length of the rod ) is [tex]L = 6.3 \ m[/tex]

     The coefficient of kinetic friction is [tex]\mu_k = 0.18[/tex]

      The kinetic frictional force is  [tex]F_k = 0.212 \ N[/tex]

     The constant speed is [tex]v = 5.1 \ m/s[/tex]

Generally the magnetic force on the rod is mathematically represented as  

      [tex]F = B * I * L[/tex]

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        [tex]F_ k = B* I * L[/tex]

=>      [tex]B = \frac{F_k}{L * I }[/tex]

=>       [tex]B = \frac{0.212}{ 6.3 * 4.1 }[/tex]

=>       [tex]B = 8.20 *10^{-3} \ T[/tex]

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

Answer:

D. Q/2

Explanation:

See attached file

Answer:

32

Explanation:

Answer:

The  force on the foundation wall is   [tex]F_f = 191394 \ N[/tex]

Explanation:

From the question we are told that

     The  depth of the hole's vertical side is  [tex]d = 2.10 \ m[/tex]

      The  width of the hole is  [tex]b = 8.90 \ m[/tex]

      The distance of the concrete wall from the front of the cellar is  [tex]c = 0.189 \ m[/tex]

Generally the area which the water from the drainage covers is mathematically represented as

        [tex]A = d * b[/tex]

substituting values

        [tex]A = 2.10 * 8.90[/tex]

       [tex]A = 18.69 \ m^2[/tex]

Now the gauge pressure exerted on the foundation wall is mathematically evaluated as

          [tex]P_g = \rho * d_{avg} * g[/tex]  

Here  is the average height foundation wall where the pressure of the water is felt and it is evaluated as

         [tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]

where [tex]h_1[/tex] at the height at bottom of the hole which is equal to  [tex]h_1 = 0[/tex]

and  [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]

        [tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]

       [tex]d_{avg} = 1.05[/tex]

Where  [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]

substituting values

          [tex]P_g = 1000 * 1.05 * 9.8[/tex]

         [tex]P_g = 10290 \ Pa[/tex]

Then the force exerted by the water on the foundation wall mathematically represented as

      [tex]F_f = P_g * A[/tex]

substituting values

      [tex]F_f = 10290 * 18.69[/tex]

       [tex]F_f = 191394 \ N[/tex]

Answer:

Inductance of a coil(L) = 0.44 H (Approx)

Explanation:

Given:

coil produces emf = 2.40 V

Old current = -27 mA

New current = 33 mA

Time taken = 11 mS

Find:

Inductance of a coil(L)

Computation:

Inductance of a coil(L) = -emf / [Δi / Δt]

Inductance of a coil(L) = -2.4 / [(-33 - 27) / 11]

Inductance of a coil(L) = -2.4 / [-5.4545]

Inductance of a coil(L) = 0.44 H (Approx)

Answer:

a)   d sin θ = m λ₀ / n

b)   d sin θ = (m + ½) λ₀ / n

c)    d = 2,439 10⁻⁷ m

Explanation:

For the interference these rays of light we must take as for some aspects,

* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2

* when the ray penetrates the lens the donut length changes by the refractive index

            λ = λ₀ / n

now let's write the destructive interference equation for these lightning bolts

           d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n

           d sin θ = m λ₀ / n

b) now nc> np

in this case there is no phase change in the reflected ray and the equation for destructive interference remains

             d sin θ = (m + ½) λ₀ / n

c) select the value of nc = 2.05 of the ZnO

we calculate the thickness of the film (d)

            d = m λ / (n sin 90)

in this type of interference the observation is normal, that is, the angle is 90º)

           d = 1 500 10-9 / (2.05 1)

           d = 2,439 10⁻⁷ m

d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference

Answer: A) [tex]P_{x}[/tex] = 564.4 N

              B) [tex]P_{y}[/tex] = 374 N

Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:

[tex]hypotenuse^{2} = side^{2} + side^{2}[/tex]

[tex]side = \sqrt{hypotenuse^{2} - side^{2}}[/tex]

side = [tex]\sqrt{3^{2} - 2.5^{2}}[/tex]

side = 1.65

Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ([tex]P_{x}[/tex]) and othe that is perpendicular ([tex]P_{y}[/tex]). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.

The image in the attachment illustrates the described above.

A) [tex]P_{x}[/tex] = P sen θ

[tex]P_{x} = P.\frac{oppositeside}{hypotenuse}[/tex]

[tex]P_{x}[/tex] = 680.[tex]\frac{1.65}{3}[/tex]

[tex]P_{x}[/tex] = 564.4 N

B) [tex]P_{y}[/tex] = P cos θ

[tex]P_{y} = P.\frac{adjacentside}{hypotenuse}[/tex]

[tex]P_{y}[/tex] = 680. [tex]\frac{1.65}{3}[/tex]

[tex]P_{y}[/tex] = 374 N

Answer:

On that line segment between the two charges, at approximately [tex]0.7\; \rm m[/tex] away from the smaller charge (the one with a magnitude of [tex]5 \times 10^{-19}\; \rm C[/tex],) and approximately [tex]1.3\; \rm m[/tex] from the larger charge (the one with a magnitude of [tex]20 \times 10^{-19}\; \rm C[/tex].)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let [tex]k[/tex] denote the Coulomb constant, and let [tex]q[/tex] denote the size of a point charge. At a distance of [tex]r[/tex] away from the charge, the electric field due to this point charge will be:

[tex]\displaystyle E = \frac{k\, q}{r^2}[/tex].

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.  

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] and [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex]. Assume that the electric field is zero at [tex]r[/tex] meters to the right of the [tex]5\times 10^{-19}\; \rm C[/tex] point charge. That would be [tex](2 - r)[/tex] meters to the left of the [tex]20 \times 10^{-19}\; \rm C[/tex] point charge. (Since this point should be between the two point charges, [tex]0 < r < 2[/tex].)

The electric field due to [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] would have a magnitude of:

[tex]\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}[/tex].

The electric field due to [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex] would have a magnitude of:

[tex]\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}[/tex].

Note that at all point in this section, the two electric fields [tex]E_1[/tex] and [tex]E_2[/tex] will be acting in opposite directions. At the point where the two electric fields balance each other precisely, [tex]| E_1 | = | E_2 |[/tex]. That's where the actual electric field is zero.

[tex]| E_1 | = | E_2 |[/tex] means that [tex]\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}[/tex].

Simplify this expression and solve for [tex]r[/tex]:

[tex]\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0[/tex].

[tex]\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0[/tex].

Either [tex]r = -2[/tex] or [tex]\displaystyle r = \frac{2}{3}\approx 0.67[/tex] will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, [tex]0 < r < 2[/tex]. Therefore, [tex](-2)[/tex] isn't a valid value for [tex]r[/tex] in this context.

As a result, the electric field is zero at the point approximately [tex]0.67\; \rm m[/tex] away the [tex]5\times 10^{-19}\; \rm C[/tex] charge, and approximately [tex]2 - 0.67 \approx 1.3\; \rm m[/tex] away from the [tex]20 \times 10^{-19}\; \rm C[/tex] charge.

Answer:

the new surface charge density = Q/4πr²( initial surface charge density divided by 4)

Explanation:

charge density(surface) = Q/A = charge/area

let r be the initial radius of the disk

therefore, area A = πr²

charge density = Q/πr²

Now that the radius is doubled, let it be represented as R

∴ R = 2r

Recall, charge density = Q/A

A = πR = π(2r)² = 4πr²

the new surface charge density = Q/4πr²

the initial surface charge density divided by 4

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

A)  The height of the balloon at the end of 40 sec is 820 feet.

B) The average velocity of the balloon is 30 ft/sec.

C) The velocity of the balloon at the end of 30 sec is

Given :

Part A)

The height of the balloon after 40 secs is :

h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

Part B)

The average velocity of the balloon is  :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0 sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

When at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon = 30 ft/sec

The average velocity of the balloon  is  30 ft/sec.

Part C)

The velocity of the balloon at the end of 30 sec is :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec.

Learn more about "Velocity":

https://brainly.com/question/862972?referrer=searchResults

Answer:

Explanation:

coefficient of linear expansion α = 2.9 x 10⁻⁵

coefficient of volume expansion γ = 3 x 2.9 x 10⁻⁵ = 8.7 x 10⁻⁵

[tex]d_t = d_0( 1 - \gamma t )[/tex]

[tex]d_{82} = 1.13\times 10^4( 1 - 8.7\times 10^{-5}\times82 )[/tex]

= 1.13 x 10⁴ - 806.14 x 10⁻¹

= 1.13 x 10⁴ - 0.00806 x 10⁴

= 1.1219 x 10⁴ kg / m³

b ) mass of the sample will remain the same as mass does not increase or decrease with temperature .

Answer:

t = 2.95 min

Explanation:

Given that,

The diameter of flywheeel, d = 1.5 m

Mass of flywheel, m = 250 kg

Initial angular velocity is 0

Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]

We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.

Firstly, we will find the angular acceleration of the flywheel.

The moment of inertia of the flywheel,

[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]

Now,

Let the torque is 50 N-m. So,

[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]

So,

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]

or

t = 2.95 min

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

[tex] M = log(\frac{I}{S}) [/tex]

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Answer:

9.1Ω

Explanation:

The circuit diagram has been attached to this response.

(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e

[tex]\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

=> [tex]R_{X} = \frac{R_1 * R_2}{R_1 + R_2}[/tex]             ------------(i)

From the question;

R1 = 3Ω,

R2 = 7Ω

Substitute these values into equation (i) as follows;

[tex]R_{X} = \frac{3 * 7}{3 + 7}[/tex]

[tex]R_{X} = \frac{21}{10}[/tex]

[tex]R_{X} = 2.1[/tex]Ω

(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.

Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e

R = Rₓ + R3

Rₓ = 2.1Ω

R3 = 7Ω

=> R = 2.1Ω + 7Ω = 9.1Ω

Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω

Answer:

In order to stay hydrated during warm-up(s) drink 8oz of water 20-30 mintues before you start exercising or during your warm-up(s), make sure you drink 7 to 10 oz of water every 10 to 20 minutes during exercise, and drink 8oz of water no more than 30 minutes after you exercise.

In order to stay hydrated during scheduled activity(s) drink 17 to 20 oz of water 2 to 3 hours before you start to exercise, like said before drink 8 oz of water 20 to 30 minutes before you start exercising or during your warm-up(s), drink 7 to 10 oz of water every 10 to 20 minutes during exercise, also said before drink 8 ounces of water no more than 30 minutes after you exercise.

Answer: My scheduled activity was one hour of softball practice. I play catcher, so my thighs and knees take a lot of abuse from kneeling and standing. The lunges were excellent at preparing my thighs for softball. The high knees exercise and arm pumping didn’t feed into softball too well. I suppose that they might help me with base running.

Explanation: EDMENTUM

Complete  Question

In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?

Answer:

a

  [tex]\theta = 0.3819^o[/tex]

b

  [tex]y = 0.00386 \ m[/tex]

Explanation:

From the question we are told that

    The slit separation is  [tex]d = 150 \lambda[/tex]

    The  distance from the screen is  [tex]D = 57.9 \ cm = 0.579 \ m[/tex]

Generally the condition for constructive interference is mathematically represented as

            [tex]dsin (\theta ) = n * \lambda[/tex]

=>        [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]

where  n is the order of the maxima  and value is 1 because we are considering the central maximum and an adjacent maximum

     and  [tex]\lambda[/tex] is the wavelength of the light

So

       [tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]

       [tex]\theta = 0.3819^o[/tex]

Generally the distance between the maxima is mathematically represented as

       [tex]y = D tan (\theta )[/tex]

=>    [tex]y = 0.579 tan (0.3819 )[/tex]

=>    [tex]y = 0.00386 \ m[/tex]

Answer:

Torque = 8.38Nm

Explanation:

Time= 8.00s

angular speed (w) =400 rpm

Moment of inertia (I)= 1.60kg.m2 about its rotation axis

We need to convert the angular speed from rpm to rad/ sec for consistency

2PI/60*n = 0.1047*409 = 41.8876 rad/sec

What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?

Then we need to use the formula below for our torque calculation

from basic equation T = J*dω/dt ...we get

Where : t= time in seconds

W= angular velocity

T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm

Therefore, constant torque that is required is 8.38 Nm

Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.

Given-

Inertia of the flywheel is 1.60 kg m squared.

Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,

[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]

[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]

[tex]\omega = 41.89[/tex]

Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.

When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,

[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]

[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]

[tex]\tau=8.38[/tex]

Hence, the required constant torque is 8.38 N-m.

to know more about the torque, follow the link below-

https://brainly.com/question/6855614

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

Answer:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Explanation:

The frequency of an electromagnetic radiation can be given by the following formula:

υ = c/λ

where,

υ = frequency of electromagnetic wave = ?

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = 290 nm to 320 nm

FOR LOWER LIMIT OF FREQUENCY:

λ = 320 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)

υ = 9.375 x 10¹⁴ Hz

FOR UPPER LIMIT OF FREQUENCY:

λ = 290 nm = 3.2 x 10⁻⁷ m

Therefore,

υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)

υ = 10.34 x 10¹⁴ Hz

Therefore, the frequency range for UVB radiations is:

υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz

Answer:

Explanation:

A microscope is made up of a convex lens and the nature of the image formed by the object viewed from it is a virtual image. Since the image is virtual, the image distance will be negative and the focal length will be positive (for convex lenses).

Using the lens formula to first calculate the image distance from the lens;

1/f = 1/u + 1/v

f is the focal length = 0.150 cm,

u is the object distance = 0.155 cm

v is the image distance.

Since the image distance is negative,

1/f = 1/u - 1/v

1/0.150 = 1/0.155 - 1/v

1/v = 1/0.155 - 1/0.15

1/v = 6.542 - 6.667

1/v = -0.125

v = 1/-0.125

v = -8 cm

Magnification = image distance/object distance

Mag = 8/0.155

Mag = 51.6

Magnification produces is 51.6

Answer:

a) 6738.27 J

b) 61.908 J

c)  [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

[tex]I[/tex] =  [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] =  [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

[tex](I_{1} +I_{2} )w[/tex]

where the subscripts 1 and 2 indicates the values first and second  flywheels

[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

where m is the mass of the car

[tex]v_{car}[/tex] is the velocity of the car

Equating the energy

2246.09 =  [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

making m the subject of the formula

mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Answer:

Explanation:

volume is same

π r₁² L₁  =π r₂²L₂

L₁ / L₂ = r₂² / r₁²

For resistance the formula is

R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area

R₁ = ρ L₁ / S₁

R₂ = ρ L₂ / S₂

Dividing

R₁ / R₂ = L₁ / L₂ x S₂ / S₁

=  L₁ / L₂ x r₂² / r²₁

= L₁ / L₂ x L₁ / L₂

= L₁²/ L₂²

L₂ = 1.1 L₁ ( GIVEN )

= L₁²/ (1.1L₁)²

1 / 1.21

R₂ / R₁ = 1.21 .

= 1.2

Answer:

The  induced current is [tex]I = 6.25*10^{-4} \ A[/tex]

Explanation:

From the question we are told that  

    The number of turns is  [tex]N = 1[/tex]

     The  cross-sectional area is  [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]

    The  initial magnetic field is  [tex]B_i = 0.500 \ T[/tex]

     The  magnetic field at time =  1.02 s  is  [tex]B_t = 2.60 \ T[/tex]

     The  resistance is  [tex]R = 2.70\ \Omega[/tex]

The  induced emf is mathematically represented as

       [tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as

        [tex]d \phi = dB * A[/tex]

Where  dB  is the change in magnetic field which is mathematically represented as

        [tex]dB = B_t - B_i[/tex]

substituting values

        [tex]dB = 2.60 - 0.500[/tex]

        [tex]dB = 2.1 \ T[/tex]

Thus  

      [tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]

     [tex]d \phi = 1.722*10^{-3} \ weber[/tex]

So  

     [tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]

     [tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]

The  induced current i mathematically represented as

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

  substituting values

       [tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]

       [tex]I = 6.25*10^{-4} \ A[/tex]

Answer:

The work done on the ball by the tension force is 0 J.

Explanation:

The work can be calculated as follows:

[tex]W = |F|\cdot |d|cos(\theta)[/tex]

Where:

F: is the tension force = 10 N

d: is the displacement = ball's circumference = 3 m

θ: is the angle between the force and the distance = 90°

Hence, the work is:

[tex]W = |10| \cdot |3| cos(90) = 0 J[/tex]

Since the tension force and the displacement vector are orthogonal, the work done on the ball is zero.

Therefore, the work done on the ball by the tension force is 0 J.

I hope it helps you!              

The work done on the ball by the 10 N tension force is zero ( 0 Joules).

Given that:

∴

Using the work equation;

W = F × d cos θ

W = 10×3× cos (90)

W = 10 × 3 × 0

W = 0 Joules

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Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

Answer:

53°

Explanation:

I/Io*2= 0.18

0.18= cos²theta

Cos^-1(0.36) = 53°

Answer:

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

While energy stored in plates capacitor is given as; U_o = Q²/2C

Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

Thus, the final energy stored between the plates of capacitor is given as;

U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)