It you balance a weight of 20 N at the 15 cm MARK and want to balance a weight of on the other side of the meter stick at 70 cm MARK, what weight should you place there

Answer:

The centripetal acceleration is [tex]a = 11.97 \ m/s^2[/tex]

Explanation:

From the question we are told that

     The radius  is [tex]r = 0.58 \ km = 0.58 * 1000 = 580 \ m[/tex]

      The speed is [tex]v = 300\ km /hr = \frac{300 *1000}{1 * 3600 } = 83.33 \ m/s[/tex]

The centripetal acceleration of the pilot is mathematically represented as

       [tex]a = \frac{v^2 }{r}[/tex]

substituting  values

      [tex]a = \frac{(83.33)^2 }{580}[/tex]

     [tex]a = 11.97 \ m/s^2[/tex]

Answer:

Yes the monkey will get hit and it will not avoid the dart.

Explanation:

Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.

No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.

Answer:

r = 0.405m = 40.5cm

Explanation:

In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]       (1)

Fc: centripetal acceleration (tension force on the string) = 12N

m: mass of the ball = 60g = 0.06kg

r: length of the string = ?

v: linear speed of the ball = 9.0m/s

You solve for r in the equation (1) and replace the values of the other parameters:

[tex]r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m[/tex]

The length of the string between Wanda and the ball is 0.405m = 40.5cm

Answer:

I can help

Explanation:

Answer:

a) The final angular velocity of the system is 4 radians per second, b) The amount of mechanical energy lost due to friction is 5.25 joules.

Explanation:

a) The problem is a clear representation of the Principle of the Angular Momentum Conservation, where moment of inertia of the system is increased by the adding of the uniform rod and there are no external forces exerted on the system. This system is represented by the following model:

[tex]I_{d} \cdot \omega_{o} = (I_{d} + I_{r})\cdot \omega_{f}[/tex]

Where:

[tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex] - Initial and final angular velocities, measured in radians per second.

[tex]I_{d}[/tex], [tex]I_{r}[/tex] - Moments of inertia of the uniform solid disk and uniform rod, measured in [tex]kg\cdot m^{2}[/tex].

Now, the final angular speed is cleared:

[tex]\omega_{f} = \frac{I_{d}}{I_{d}+I_{r}}\cdot \omega_{o}[/tex]

The moments of inertia of the uniform solid disk and uniform rod are modelled by these formulas:

Solid Disk

[tex]I_{d} = \frac{1}{2}\cdot m_{d}\cdot r_{d}^{2}[/tex]

Where:

[tex]m_{d}[/tex] - Mass of the disk, measured in kilograms.

[tex]r_{d}[/tex] - Radius of the disk, measured in meters.

Given that [tex]m_{d} = 1\,kg[/tex] and [tex]r_{d} = 1\,m[/tex], the moment of inertia of the disk is:

[tex]I_{d} = \frac{1}{2}\cdot (1\,kg)\cdot (1\,m)^{2}[/tex]

[tex]I_{d} = 0.5\,kg\cdot m^{2}[/tex]

Rod (rotating about its center)

[tex]I_{r} = \frac{1}{12}\cdot m_{r}\cdot l_{r}^{2}[/tex]

Where:

[tex]m_{r}[/tex] - Mass of the rod, measured in kilograms.

[tex]l_{r}[/tex] - Length of the rod, measured in meters.

Given that [tex]m_{r} = 0.5\,kg[/tex] and [tex]l_{r} = 3\,m[/tex], the moment of inertia of the rod is:

[tex]I_{r} = \frac{1}{12}\cdot (0.5\,kg)\cdot (3\,m)^{2}[/tex]

[tex]I_{r} = 0.375\,kg\cdot m^{2}[/tex]

Now, knowing that [tex]\omega_{o} = 7\,\frac{rad}{s}[/tex], the final angular velocity is:

[tex]\omega_{f} = \left(\frac{0.5\,kg\cdot m^{2}}{0.5\,kg\cdot m^{2}+0.375\,kg\cdot m^{2}}\right)\cdot \left(7\,\frac{rad}{s} \right)[/tex]

[tex]\omega_{f} = 4\,\frac{rad}{s}[/tex]

The final angular velocity of the system is 4 radians per second.

b) According to the Principle of Energy Conservation, the inclusion of the uniform rod on the turntable is represented by the following expression:

[tex]K_{1} = K_{2} + \Delta E_{loss}[/tex]

Where:

[tex]K_{1}[/tex] - Rotational kinetic energy of the uniform disk, measured in joules.

[tex]K_{2}[/tex] - Rotational kinetic energy of the system (uniform disk + uniform rod), measured in joules.

[tex]\Delta E_{loss}[/tex] - Mechanical energy lost due to friction, measured in joules.

The mechanical energy lost due to friction is cleared:

[tex]\Delta E_{loss} = K_{1} - K_{2}[/tex]

Now, the expression is expanded and mechanical energy losses is calculated:

[tex]\Delta E_{loss} = \frac{1}{2}\cdot I_{d}\cdot \omega_{o}^{2} - \frac{1}{2}\cdot (I_{d}+I_{r})\cdot \omega_{f}^{2}[/tex]

[tex]\Delta E_{loss} = \frac{1}{2}\cdot (0.5\,kg\cdot m^{2})\cdot \left(7\,\frac{rad}{s} \right)^{2} - \frac{1}{2}\cdot (0.5\,kg\cdot m^{2} + 0.375\,kg\cdot m^{2})\cdot \left(4\,\frac{rad}{s} \right)^{2}[/tex]

[tex]\Delta E_{loss} = 5.25\,J[/tex]

The amount of mechanical energy lost due to friction is 5.25 joules.

Question:

Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.652 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer in Joules (J)

Answer:

-263J

Explanation:

Though its difficult and infact impossible to measure the internal energy of a system, the change in internal energy ΔE, can however be determined. This change when it is accompanied by work(W) and transfer of heat(Q) in or out of the system, can be calculated as follows;

ΔE = Q + W       ----------------(i)

Q is negative if heat is lost. It is positive otherwise

W is negative if work is done by the system. It is positive otherwise.

From the question;

Q = -0.652kJ = -652J    {the negative sign shows heat loss}

W = +389J                      {the positive sign shows work done on the system(balloon)}

Substitute these values into equation (i) as follows;

ΔE = -652 + 389

ΔE = -263J

Therefore the change in internal energy is -263J

PS: The negative sign shows that the process is exothermic. This means that the system (balloon) lost some energy to the environment.

Answer:

The  acceleration is [tex]a = 5 \ m/s^2[/tex]  

Explanation:

From the  question we are told that

      The  coefficient of kinetic friction is  [tex]\mu_k = 0.24[/tex]

       The coefficient of static friction is  [tex]\mu_s = 0.75[/tex]

       The horizontal force is [tex]F_h = 93 \ N[/tex]

Generally the static frictional force is  mathematically represented as

         [tex]F_F = \mu_s * (m * g )[/tex]

The  static frictional force is the equivalent to the maximum possible force for which the crate does not begin to slide So

       [tex]F_h = F_F = \mu_s * (m * g )[/tex]

=>      [tex]93 = \mu_s * (m * g )[/tex]

=>        [tex]m = \frac{93}{\mu_s * g }[/tex]

substituting values  

          [tex]m = \frac{93}{0.75 * 9.8 }[/tex]

        [tex]m = 12.65 \ kg[/tex]

When the crate is already sliding the frictional force is

      [tex]F_s = \mu_k *(m * g )[/tex]

substituting values  

     [tex]F_s = 0.24 * 12.65 * 9.8[/tex]

     [tex]F_s = 29.82 \ N[/tex]

Now the net force when the horizontal force is applied during sliding is  

      [tex]F_{net} = F_h - F_s[/tex]

substituting values  

     [tex]F_{net} = 93 - 29.8[/tex]

     [tex]F_{net} = 63.2 \ N[/tex]

This  net force is mathematically represented as

     [tex]F_{net } = m * a[/tex]

Where a is the acceleration of the crate

So  

      [tex]a = \frac{F_{net}}{m }[/tex]

      [tex]a = \frac{ 63.2}{12.65 }[/tex]

      [tex]a = 5 \ m/s^2[/tex]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

Answer:

The current in the coil is 4.086 A

Explanation:

Given;

radius of the circular coil, R = 2.5 cm = 0.025 m

number of turns of the circular coil, N = 740 turns

magnetic field at the center of the coil, B = 0.076 T

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

[tex]B = \frac{N\mu_o I}{2R}[/tex]

where;

μ₀ is permeability of free space = 4 x 10⁻⁷ m/A

I is the current in the coil

R is radius of the coil

N is the number of turns of the coil

The current in the circular coil is given by

[tex]B = \frac{N\mu_o I}{2R} \\\\I = \frac{2BR}{N\mu_o} \\\\I =\frac{2*0.076*0.025}{740*4\pi*10^{-7}} \\\\I = 4.086 \ A[/tex]

Therefore, the current in the coil is 4.086 A

Answer:

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

Explanation:

- To find the direction of the conventional current in the wire you use the following formula:

[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex]       (1)

i: current in the wire = ?

F: magnitude of the magnetic force on the wire = 15.1N

B: magnitude of the magnetic field = 6.1T

l: length of the wire that is affected by the magnetic field = 0.45m

The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).

The direction of the current must be in the +y direction (+^j). In fact, you have:

^j X ^k = ^i

The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):

[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

Answer:

If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.

Explanation:

The frequency of a vibrating string is primarily based on three factors:

The sounding length (longer is lower, shorter is higher)

The tension on the string (more tension is higher, less is lower)

The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)

To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.

Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.

And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.

Many things can influence the amplitude.

What is producing the sound?

How far are you from the source of the sound? The farther away the smaller the amplitude.

Intervening material. Sound does not travel through walls as well as air.

Depends on what is detecting the wave sound. Ear vs. microphone.

Answer:

The frequency will determine the pitch

the amplitude will determine the loudness

Explanation:

The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.

The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.

Answer:maximum

Explanation:

Answer:

D) 3.0

Explanation:

As we know that

[tex]The\ specific\ gravity\ of\ fluid = \frac{density\ of\ fluid}{density\ of\ water \ at\ 4\ C} = \frac{\rho_f}{1,000}[/tex]

[tex]0.6 = \frac{\rho_f}{1000}[/tex]

So,

[tex]\rho_f = 600[/tex]

Now

T = True weight of object

= mg

[tex]= 25 \times 9.8[/tex]

= 245 N

W = apparent weight = 200 N

[tex]\sigma[/tex] = density of object

Now we use the formula

buoyancy force = True weight - W

[tex]\rho_f V g = 245 - 200[/tex]

600 V (9.8) = 45

V = 0.007653

m = 25 kg

And as we know that

[tex]\sigma = \frac{m}{V}[/tex]

[tex]= \frac{25}{0.007653}[/tex]

= 3266.7

Now  

specific gravity is

[tex]= \frac{\sigma}{water\ density}[/tex]

[tex]= \frac{3266.7}{1000}[/tex]

= 3.2  

Hence, the correct option is d.

Answer:

Explanation:

The distance of middle point from centres of spheres will be as follows

From each of 2 cm diameter sphere

R  = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m

Expression for electric field = Q / 4πε R²

Electric field due to positive charge

E₁ = 70  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 33.3 x 10⁴ N/C

Electric field due to negative  charge

E₂ = 40  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 19.02 x 10⁴ N/C

E₁ and E₂ act in the same direction so

Total field = (33.3 + 19.02 ) x 10⁴

= 52.32 x 10⁴ N/C .

Answer:

100% efficiency

Explanation:

the machine has _100%_ efficiency.

(fill in the blank)

Answer:

100%

Explanation: A p e x

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.

The position of a particle is r(t)= (4.0t²i+ 2.4j- 5.6tk) m. (Express your answers in vector form.)

(a) Determine its velocity (in m/s) and acceleration (in m/s²) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.)

v(t)= ________m/s

a(t)= ________m/s²

(b) What are its velocity (in m/s) and acceleration (in m/s²) at time t 0?

v(0) =_______ m/s

a(0)=_______ m/s²

(a)

v(t)= [tex]8ti - 5.6k[/tex] m/s

a(t)= 8i m/s²

(b)

v(0) = -5.6k m/s

a(0)= 8i m/s²

From the question, the position of the particle is given by;

r(t)= (4.0t²i+ 2.4j- 5.6tk)        -----------------(i)

(a)

(i)To get the velocity, v(t), of the particle, we'll take the first derivative of the position of the particle (given by equation (i)) with respect to time, t, as follows;

v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]\frac{d(4.0t^2i + 2.4j - 5.6tk)}{dt}[/tex]

v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]8ti +0j - 5.6k[/tex]

v(t) = [tex]8ti - 5.6k[/tex]          --------------------(ii)

(ii) To get the acceleration, a(t), of the particle, we'll take the first derivative of the velocity of the particle (given by equation (ii)) with respect to time, t, as follows;

a(t) = [tex]\frac{dv(t)}{dt}[/tex]  = [tex]\frac{d(8ti - 5.6k)}{dt}[/tex]

a(t) = 8i                    --------------------(iii)

(b)

(i) To get the velocity of the particle at time t = 0, substitute the value of t = 0 into equation (ii) as follows;

v(t) = [tex]8ti - 5.6k[/tex]  

v(0) = 8(0)i - 5.6k

v(0) = 0 - 5.6k

v(0) = -5.6k

(ii) To get the acceleration of the particle at time t = 0, substitute the value of t = 0 into equation (iii) as follows;

a(t) = 8i

a(0) = 8i

Explanation:

A projectile is usually launched by an initial force, is influenced by gravity forces and air resistance, and takes a parabolic, or more accurately, an  elliptical path. At the beginning of the launch of a rocket, the rocket is mostly under the effect of its engine thrust, and takes a vertical flight path upwards; using engine thrust to maneuver itself to remain in this vertical flight till it is almost at its required orbit. At this stage, the rocket can't be said to follow a projectile path. When the rocket gets to the pitchover stage, at which it is almost ready to enter its atmosphere, the engine thrust is maneuvered to turn the rocket by an angle of attack. At this stage, the flight path is no longer vertical. At this point after pitchover, in which the flight path is no longer vertical, gravity tends to pull the rocket down in a parabolic path. This is the point where the rocket acts as a projectile, but is countered by engine thrust.

Answer:

a

Explanation:

Answer:

P= 2414.9 Pa

Explanation:

given

density of air , p = 1.29 kg/m³

speed of air over the upper surface , v₁ = 125 m/s

speed of air over the lower surface , v₂ = 109 m/s

the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)

P = 0.5 × 1.29 × ( 125² - 109²)

P= 0.645(3744)

P = 2414.9 Pa

the pressure difference on an airplane wing is 2414.9 Pa

Answer:

The test charge will take the south-west direction indicated in option 6.

Explanation:

The image is shown below.

Since all the charges are positively charged, they will all repel each other. If we consider the force on +q due to +Q and +Q, then we can proceed as follows

The +Q particle at the top left corner of the cube will exert a vertical downward force on +q in the -ve y-axis.

The +Q particle at the bottom right corner of the cube will exert a force on +q towards the horizontal left on the -ve x-axis.

Both of these forces will act at angle of 90°, and therefore, the resultant force will act at an angle of 45° to horizontal and vertical forces.

The result is that the +q charge will move in a south-west direction of the cube.

Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.

Explanation:

The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:

F = BqVSinØ

If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field

According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.

Answer:

The  potential is  [tex]V = 153.659 \ V[/tex]

Explanation:

From the question we are told that

     The diameter of the plastic sphere is  [tex]d = 0.410 \ cm = 0.0041 \ m[/tex]

      The magnitude of the charge is  [tex]q = 35.0 pC = 35.0 *10^{-12} \ C[/tex]

The radius of the plastic sphere is  mathematically evaluated as

          [tex]r = \frac{d}{2}[/tex]

=>     [tex]r = \frac{0.0041}{2}[/tex]

       [tex]r = 0.00205 \ m[/tex]

The  potential near the surface is mathematically represented as

         [tex]V = \frac{k * q}{r }[/tex]

Where k is the Coulombs constant with value [tex]9 *10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

substituting values  

       [tex]V = \frac{9*10^9 * 35 *10^{-12}}{0.00205}[/tex]

       [tex]V = 153.659 \ V[/tex]

Answer:

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Explanation:

The power dissipation by an electrical circuit is given by the following formula:

Power Dissipation = (Voltage)(Current)

P = VI

but, from Ohm's Law, we know that:

Voltage = (Current)(Resistance)

V = IR

Substituting this in formula of power:

P = (IR)(I)

P = I²R   ---------------- equation 1

Now, if we double the current , then the power dissipated by that circuit will be:

P' = I'²R

where,

I' = 2 I

Therefore,

P' = (2 I)²R

P' = 4 I²R

using equation 1

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Answer:

From the initial height h

Explanation:

When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already

What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.

28.29m/s

In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;

([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v            -----------------(i)

Where;

[tex]m_{C}[/tex] = mass of car

[tex]u_{C}[/tex] = initial velocity of car before collision

[tex]m_{D}[/tex] = mass of deer

[tex]u_{D}[/tex] = initial velocity of the deer before collision

v = common velocity with which the car and the deer move after collision

From the question;

[tex]m_{C}[/tex] = 900kg

[tex]u_{C}[/tex] = +30.0m/s    (direction of the motion of the car taken positive)

[tex]m_{D}[/tex] = 150kg

[tex]u_{D}[/tex] = +18.0m/s    (relative to the direction of the car, the velocity of the deer is also positive )

Substitute these values into equation (i) as follows;

(900 x 30.0) + (150 x 18.0) = (900 + 150)v

27000 + 2700 = 1050v

29700 = 1050v

v = [tex]\frac{29700}{1050}[/tex]

v = 28.29m/s

Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.

Answer:

The correct answer is "Resistor".

Explanation:

So that the above seems to be the correct solution.

Answer:

Pls see attached file

Explanation:

The electric flux through annular ring is parallel to the surface everywhere hence the angle of filled lines will be π/2 and thus cosine of this angle is zero leads to the electric flux Φ3 = 0.

Electric flux the flow of electric field lines that passing over a given area in in unit time. This is actually the field line density in a surface. This physical quantity is dependent on the magnitude of field, radius of the object if it is a ring and the charge.

Let the area of an infinitesimal surface be dA and the field acting is E then flux is the dot product E(r) .dA.

The field respect to a position r for the radius r1 is written as follows:

E(r) = (C/r² )

where, c is a proportionality constant for r.

The integrand equation for the electric flux is written as follows:

Ф3= E(r).dA = E(r).dA cos ∅

Consider the surface 3 in the annular ring where dA is normal to the field E(r) and the electric field is parallel to everywhere in the surface so the angle will be  π/2. Thus ,cos  π/2 is zeo making Ф3.

To find more about electric flux, refer the link below:

https://brainly.com/question/14850656

#SPJ5

Your question is incomplete. But your complete question includes the image attached with the answer.

Answer:

t = 1.098*RC

Explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

[tex]Q=Q_{max}[1-e^{-\frac{t}{RC}}][/tex]         (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax    

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

[tex]Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC[/tex]

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC

Period = 1 / frequency

Period = 1 / (35,621 /s)

Period = 2.8073... x 10⁻⁵ sec

Period = 28.07 microseconds

or

Period = 0.0281 millisecond

The period (T) of the crystal's motion is equal to [tex]2.81 \times 10^{-5}\;seconds[/tex].

Given the following data:

To calculate the period of the crystal's motion:

Mathematically, the period of oscillation of an object is given by this formula:

[tex]T=\frac{1}{F}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]T=\frac{1}{35621} \\\\T=2.81 \times 10^{-5}\;seconds[/tex]

Therefore, the period (T) of the crystal's motion is equal to [tex]2.81 \times 10^{-5}\;seconds[/tex].

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