The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of its full equilibrium charge

Answer:

a) using

R=V/I =117/12.3

R=9.5 ohms

b)

E=V-I*R =117-3.38*9.5

E=84.8Volts

c)

at (1/3)rd of normal speed ,back emf is (1/3) of its maximum

value

E=(1/3)*84.8=28.3Volts

Current drawn

I=V-Eback/R =117-28.3/9.5

I=9.33A

Explanation:

The resistance is = 9.5 ohms

The back emf developed at normal speed is = 84.90 volts

The current drawn at one-third normal speed =9.33 A.

To calculate the resistance of the armature coil this formula is used;

V = IR

make R the subject of formula,

R = V/I

where R = resistance

V = voltage

I = Current

R = 117/12.3

R = 9.5 ohms

To calculate the back emf developed at normal speed, this formula is used;

E = V + Ir ( for normal emf)

But for back emf which is the difference between the supplied voltage and the loss from the current through the resistance, this formula is used;

E = V - Ir

where V = 117v

I = 3.38

r = 9.5

E = 117 - ( 3.38 × 9.5)

= 117 - 32.11

= 84.90 volts

To calculate the current drawn at one-third normal speed;

1/3 of Emf = 1/3 × 84.90

= 28.3volts

Therefore current (I) = V - E/ R

= 117- 28.3/9.5

= 88.7/9.5

= 9.33 A

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Answer:

The double slit experiment showed for the first time that light can be interfered, producing bands of light and dark fringes on a screen. This phenomenon was a unique and typical characteristic of waves.

Explanation:

Th double slit experiment by Thomas Young proved, and sealed for the first time the wave nature of light; showing that light just as any other wave can produce interference which was a unique, typical phenomenon of waves. The Interference of light was shown by allowing light to pass through narrow slits and superimpose on a wall or screen, at a distance away from the slit, producing a clear pattern of light and dark fringes. This was the first experiment to proof that darkness can be produced by the addition of light on light. Interference is accompanied by a spatial redistribution of the optical intensity without violation of power conservation. The phenomenon of interference proved the intuitive ideas of Huygens regarding the wave nature of light against Newton's particle nature of light theory.

Answer:

I believe it's called rapid growth

Explanation:

that is my answer no matter what

Explanation:

proton number + neutron number = atomic mass

30 + 35 = 65

Answer:

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     0------5cm-----------------15cm-------------P---------30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium

Answer:

Anti-Particle

Explanation:

The statement "our sun is an 'average' sun" is  true when it is used to describe or characterize some unique physical properties of stars generally in the universe. 'Average' in this sense is used to define a typical sun such as, "stars should glow like our sun an average star."

The statement is used wrongly when used to in quantifying other stars in the universe, based on calculated values from our sun. In this case, we cannot truly say if our sun is a true representative average of other stars in the universe.

Yes! it is useful to characterize the milky way by simply citing the average sun. Properties like their ability to glow and radiate heat can be defined by citing an average star like our sun, so long as we don't translate it into citing quantitative properties of the sun as an average of our Milky Way Galaxy like the mass, temperature, etc.

Answer:

Yes the glasses will be fogged

Explanation:

See attacher file

Answer:

6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex]

Explanation:

From Wien's displacement formula;

Q = e A[tex]T^{4}[/tex]

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = [tex]\frac{Q}{A}[/tex] = e[tex]T^{4}[/tex]

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × [tex](1000)^{4}[/tex]

                             = 0.6 × 1.0 × [tex]10^{12}[/tex]

                             = 6.0 × [tex]10^{11}[/tex] [tex]\frac{W}{m^{2} }[/tex]

Therefore, the emissive intensity coming out of the surface is 6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex].

Answer:

R = 42.67 ohms

Explanation:

It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.

It means that when two batteries are connected in series, then the total voltage is 3.2 volts

Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :

[tex]V=IR\\\\R=\dfrac{V}{I}[/tex]

So,

[tex]R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega[/tex]

So, the resistance of the bulb is 42.67 ohms.

Answer:

[tex]1.33\times 10^{-8} N[/tex]

Explanation:

According to the given scenario, the computation of force between them is shown below:-

As we know that

[tex]\mu = 4\pi\times 10^{-7}[/tex]

The force between two current carrying wires will be

[tex]F = \frac{\mu_oI_1I_2L}{2\pi r}[/tex]

[tex]= \frac{4\pi\times 10^{-7} (1 A) (1 A) (1.0 m)}{2\pi (1.5m)}[/tex]

[tex]= 1.33\times 10^{-7} N[/tex]

[tex]= 1.33\times 10^{-8} N[/tex]

Therefore for computing the force between two wires we simply applied the above formula.

So, the force between two wires carrying 1 A current [tex]= 1.33\times 10^{-8} N[/tex].

Answer:

  B : is independent of the natural frequency of the oscillator

Explanation:

You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.

The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.

Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

A) The speed of the scooter at which the driver will feel weightlessness is;

v = 68.586 m/s

B) The apparent weight of both the driver and the scooter at the top of the hill is;

F_net = 779.1 N

We are given;

Mass; m = 159 kg

Radius; r = 480 m

A) Since it's motion about a circular hill, it means we are dealing with centripetal force.

Formula for centripetal force is given as;

F = mv²/r

Now, we want to find the speed of the scooter if the driver feels weightlessness.

This means that the centripetal force would be equal to the gravitational force.

Thus;

mg = mv²/r

m will cancel out to give;

v²/r = g

v² = gr

v = √(gr)

v = √(9.8 × 480)

v = √4704

v = 68.586 m/s

B) Now, he is travelling with speed of;

v = 68.586 m/s

And the radius is 2r

Let's first find the centripetal acceleration from the formula; α = v²/r

Thus; α = 4704/(2 × 480)

α = 4.9 m/s²

Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;

F_app = m(g - α)

F_net = 159(9.8 - 4.9)

F_net = 779.1 N

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Answer:

9.82 rads/sec

Explanation:

We are given;

Time taken; t = 1.6 secs

Number of somersaults = 2

Now, we know that,

1 revolution = 2π radians

And number of somersaults is the same thing as number of revolutions

So,

Total radians = 2π × 2 = 5π

Angular velocity = total number of revolutions/time period = 5π/1.6 = 9.82 rads/sec

Answer:

The answer is B. When the magnet is placed on a globe to correctly align with Earth’s magnetic field, it is considered to be suspended freely. The Earth has geographical poles as well with North and South poles. Since unlike poles attract, the South Pole of the magnet will be attracted to the geographical North.

Explanation:

B)Place the magnet vertically on the equator, with the south end facing the North pole.

A bar magnet is a square piece of an item, made from iron, metal, or every other ferromagnetic substance or ferromagnetic composite, that indicates everlasting magnetic homes. It has two poles, a north and a south pole such that when suspended freely, the magnet aligns itself so that the northern pole factors towards the magnetic north pole of the earth.

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#SPJ2

a)The heat transfer coefficient  will be 0.231 .The concept of the Newtons law of cooling is used in the given problem.

The pace at which an item cools is proportional to the temperature differential between the object and its surroundings,

According to Newton's law of cooling. Simply explained, in a cold environment, a glass of hot water will cool down faster than in a hot room.

The given data in the problem is;

T₀ is the initial temperature= 40°

[tex]\rm T_S[/tex]  is the constant temperature = 70°

T is the final temperature = 55°

From the given equation the Newtons law of cooling;

[tex]\rm T=T_S+(T_0-T_S)e^{-kt} \\\\ 55=70+(40-70)e^{-3k} \\\\-30 e^{-3k}=-15\\\\ e^{-3k}= -\frac{1}{2} \\\\ ln e^[3k}=ln 2 \\\\ K=\frac{1}{3} ln2 \\\\ K=0.231[/tex]

Hence the heat transfer coefficient  will be 0.231 .

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Answer:

Explanation:

area of the coil  A = .08 x .08 = 64 x 10⁻⁴ m ²

flux through the coil Φ = area of coil x no of turns x magnetic field

= 64 x 10⁻⁴ x 50 x B where B is magnetic field

emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2

= 1.6 B

current induced = emf induced / resistance

12 x 10⁻³ = 1.6 B / 15

B = 112.5 x 10⁻³ T .

Answer:

Explanation:

Potential energy stored in the spring = 25 J

This energy is converted into gravitational potential energy . If h be the height attained

gravitational potential energy = mgh

mgh = 25

.15 x 9.8 x h = 25

h = 17 m

Answer:

Pressure is  a scalar quantity defined as per unit area.

Density is the objects ,times its  the acceleration due to gravity.

Explanation:

Pressure is the alternative object increases the area of contact decrease .

Pressure is the force component  to the surface used to calculate pressure.

pressure is that collisions of the gas to container as the per unit time .

pressure is an physical important quantity to play the solid and  fluid .

Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.

Density is the  objects, times, volume of the object that times acceleration objects.

Density is the used to the system complex objects and materials.

Density  force is the weight of a region or objects static fluid.

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

k = 17043.5 N/m = 17.04 KN/m

Answer:

[tex]F_x[/tex] = -1.5t

[tex]F_y[/tex] = 0.25

[tex]F_{z}[/tex] = 0

Explanation:

Given equation;

p = [(-0.75 kgm/s³)t² + (3.0 kgm/s)] i + (0.25 kgm/s²)t j.

From Newton's law, the rate of change of momentum of a body is the net force acting on that body. i.e

∑F = [tex]\frac{dp}{dt}[/tex]       -----------(i)

Substitute the equation of p into equation (i) and differentiate with respect to t as follows;

∑F = [tex]\frac{dp}{dt}[/tex] = [tex]\frac{d| [(-0.75)t^{2} + (3.0)] i + (0.25)t j|}{dt}[/tex]

∑F = [tex]\frac{dp}{dt}[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex]

∑F = [tex][-1.5t + 0]i + 0.25j[/tex]

But

∑F = [tex]F_xi + F_yj + F_zk[/tex]

Where;

[tex]F_x, F_y, F_z[/tex] are the components of the net force in the x, y and z direction respectively.

=> [tex]F_xi + F_yj + F_zk[/tex] =  [tex][-1.5t + 0]i + 0.25j[/tex] = [tex]-1.5ti + 0.25j[/tex]

=> [tex]F_x[/tex] = -1.5t

=> [tex]F_y[/tex] = 0.25

=> [tex]F_{z}[/tex] = 0

Answer:

Explanation:

We shall find electric field at origin due to two given charges sitting   on the either side of origin .

Total field will add up due to their same direction .

Field due to a charge Q

= 9 x 10⁹ x Q / R²  ;  R is distance of point , Q is charge

Field due to first charge

= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴

= 90 x 10¹⁰ N/C

Field due to second  charge

= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴

= 112.5 x 10¹⁰ N/C

Total field

= 202.5 x 10¹⁰ N/C

Force on given charge at origin

= charge x field

= 4 x 10⁻³ x 202.5 x 10¹⁰

= 810 x 10⁷ N .

Answer:

4.1 N/C

Explanation:

First of all, we know from maths that the surface area of a sphere = 4πr²

Charge on inner sphere ..

Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²

Q(i) = 5.03*10^-14 C

Charge on outer sphere

Q(o) = 60*10^-12 x 4π(0.03m)²

Q(o) = 6.79*10^-13 C

Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.

For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge

= +6.79*10^-13C + (+5.03*10^-14C)

= +7.29*10^-13 C

Now again, we have

E = kQ /r²

E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²

E = 6.561*10^-3 / 1.6*10^-3

E = 4.10 N/C

Thus, the magnitude of the electric field is 4.1 N/C

Answer:

Explanation:

distance of third dark fringe

= 2.5 x λ D / d

where λ is wavelength of light , D is screen distance and d is slit separation

putting the given values

required distance = 2.5  x 739 x 10⁻⁹  x 6.3 / .49 x 10⁻³

= 23753.57 x 10⁻⁶

= 23.754 x 10⁻³ m

= 23.754 mm .

Answer:

When reviewing the correct answer is A

Explanation:

The magnetic force is given by the expression

           F = qv xB

where the bold letters indicate vectors, from this expression the module can be calculated

          F = = q v b sin θ

the direction of the force is given by the rule of the right hand, for a positive charge the speed held by the thumb, the extended fingers point in the direction of the magnetic field and the palm points the direction of the force

in this case

the speed is in the negative part of the z axis

the force is in the negative direction of the axis and

consequently the magnetizing field is in the positive direction of the x axis

When reviewing the correct answer is A

Answer:

0.3nanocouloumb

Explanation:

Pls see attached file

Answer:

  x_total = 0.17m

Explanation:

We can treat this exercise with the kinematics equations, where in the first part it is accelerated and in the second it is a uniform movement.

Let's analyze accelerated motion

The time that lasts is t = 20 10⁻³ s, the initial speed is zero (v₀ = 0), let's find the length that advances

            x₁ = v₀ t + ½ a t²

            x₁ = ½ a t²

            x₁ = ½ 210 (20 10⁻³)²

            x₁ = 4.2 10⁻² m

let's find the speed for the end of this movement

            v = v₀ + a t

            v = 0 + 210   20 10⁻³

            v = 4.2 m / s

with this speed we can find the distance that the uniform movement

           x₂ = v t2

           x₂ = 4.2   30 10⁻³

           x₂ = 1.26 10⁻¹ m

           x₂ = 0.126m

the total distance traveled is

          x_total = x₁ + x₂

          x_total = 0.0420 +0.126

          x_total = 0.168m

     Let's reduce the significant figures to two

          x_total = 0.17m

Answer:

Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely

used integrated circuits for creating clock pulses is called a 555 timer.  shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?

ANSWER : R1 = 144.3Ω,   R2 =  72.2 Ω

Explanation:

Frequency = 10 MHz

Time period = 1 / F =  0.1 u s

Duty cycle = 75% = 0.75

Duty cycle can be represented as :   Ton / T

Also: Ton = Th = 0.75 * 0.1 u s  = 75 n s

TL = T - Th = 100 ns - 75 n s = 25 n s

To find the value of R2 we use the equation for  time spent in the low (0 V) state

TL = R2*C*ln(2)

hence R2 = TL / ( C * In 2 )

c = 500 pF

Hence R2 = 25 / ( 500 pF * 0.693 )  = 72.2 Ω

To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,

Th = (R1 + R2)*C*ln(2)

  from the equation make R1 the subject of the formula

R1 =  (Th - ( R2 * C * In2 )) / (C * In 2)

R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )

R1 = ( 75 ns  - ( 25 ns ) / 500 pf * 0.693

     = 144.3Ω

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 [tex]\theta _d =0.19 ^o[/tex]

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  [tex]n_F = 1.497[/tex]

     The  wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]

    The  refractive index of the transparent acrylic plastic for red light is  [tex]n_C = 1.488[/tex]

       The  wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]

    The incidence angle is  [tex]i = 45^o[/tex]

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       [tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]

Where  [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]

So

     [tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]

      [tex]r_F = 28.18^o[/tex]

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       [tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]

Where  [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]

So

     [tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]

      [tex]r_F = 28.37^o[/tex]

The angle between the blue beam and the red beam in the acrylic block

     [tex]\theta _d = r_C - r_F[/tex]

substituting values

       [tex]\theta _d = 28.37 - 28.18[/tex]

       [tex]\theta _d =0.19 ^o[/tex]

Answer:

The vertical velocity of the skater upon landing is 10.788 meters per second.

Explanation:

Skateboarder experiments a parabolic movement. As skateboarder jumps horizontally off the top of the staircase, it means that vertical component of initial velocity is zero and accelerates by gravity, the final vertical speed is calculated by the following expression:

[tex]v = v_{o} + g\cdot t[/tex]

Where:

[tex]v_{o}[/tex] - Initial vertical speed, measured in meters per second.

[tex]v[/tex] - Final vertical speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

Given that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 1.10\,s[/tex], the final velocity of the skater upon landing is:

[tex]v = 0\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (1.10\,s)[/tex]

[tex]v = -10.788\,\frac{m}{s}[/tex]

The vertical velocity of the skater upon landing is 10.788 meters per second.

Answer:

-10.8

Explanation:

Answer on Khan Academy