An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected to a battery and charged until its plates carry charges Q and - Q, and then disconnected from the battery. If the separation between the plates is now doubled, the potential difference between the plates will

Answer:

standard kilogram is the SI unit of mass

Answer:

The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]

Force:

[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]

Answer:

13.3 pounds.

Explanation:

For a spring of constant K, with an attached object of mass M, the period can be written as:

T = 2*π*√(M/K)

Where π = 3.14

First, we know that the period is 4 seconds, then we have:

4s = (2*π)*√(M/K)

We know that if the mass is reduced by 10lb, the period becomes 2s.

Then the new mass of the object will be: (M - 10lb)

Then the period equation becomes:

2s = (2*π)*√((M-10lb)/K)

So we have two equations:

4s = (2*π)*√(M/K)

2s = (2*π)*√((M-10lb)/K)

We want to solve this for M.

First, we need to isolate K in one of the equations.

Let's isolate K in the first one:

4s = (2*π)*√(M/K)

(4s/2*π) = √(M/K)

(2s/π)^2 = M/K

K = M/(2s/π)^2 = M*(π/2s)^2

Now we can replace it in the other equation.

2s = (2*π)*√((M-10lb)/K)

First, let's simplify the equation:

2s/(2*π) = √((M-10lb)/K)

1s/π =  √((M-10lb)/K)

(1s/π)^2 =  ((M-10lb)/K

K*(1s/π)^2 = M - 10lb

Now we can use the equation: K =  M*(π/2s)^2

then we get:

K*(1s/π)^2 = M - 10lb

(M*(π/2s)^2)*(1s/π)^2 = M - 10lb

M/4 = M - 10lb

10lb = M - M/4

10lb = (3/4)*M

10lb*(4/3) = M

13.3 lb = M

Answer:

a) fem = 5.709 V,  b)  t = 0.196 s,  c)  t = 0.589 s, d)   T = 0.785 s

Explanation:

This is an exercise in Faraday's law

          fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]

          fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]

The magnetic field and the area are constant

          fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]

          fem = - N B A (-sin θ)  [tex]\frac{d \theta}{dt}[/tex]

          fem = N B (π d² / 4) sin θ   w

          fem= [tex]\frac{\pi }{4}[/tex]  N B d² w sin θ

with this expression we can correspond the questions

a) the peak of the electromotive force

this hen the sine of the angle is 1

         sin θ = 1

         fem = [tex]\frac{\pi }{4}[/tex]   77  1.18  0.10² 8.0

         fem = 5.709 V

b) as the system has a constant angular velocity, we can use the angular kinematics relations

          θ = w₀ t

          t = θ/w₀

Recall that the angles are in radians, so the angle for the maximum of the sine is

           θ= π/2

           t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]

           t = 0.196 s

c) for the electromotive force to be negative, the sine function of being

            sin θ= -1

whereby

          θ = 3π/ 2

          t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]  

          t = 0.589 s

d) This electromotive force has values ​​that change sinusoidally with an angular velocity of

          w = 8 rad / s

angular velocity and period are related

          w = 2π / T

          T = 2π / w

          T = 2π / 8

          T = 0.785 s

Answer:

nuclear fusion

Answer:

nuclear fusion

Explanation:

The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.

Answer:

the person's heart rate

Answer:

the required solution is; x(t) = 0.675sin( 2.222t )

Explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5

given that mass m = 9 kg

[tex]x[/tex] = 1.8 m

k is F / x

hence

k = F / x

given that, F = 80 N

we substitute

k = 80 / 1.8

k = 44.44

so

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,

we input

9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,

[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0

so auxiliary equation is,

r² + 4.9377 = 0

r² = -4.9377

r = √-4.9377

r = ±2.222i

hence, the solution will  be;

x(t) = A×cos( 2.222t ) + B×sin( 2.222t )

⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

[tex]x^t[/tex](t) = 1.5

1.5 = 2.222B

so

B = 1.5 / 2.222 = 0.675

Hence, the required solution is; x(t) = 0.675sin( 2.222t )

Answer:

The first person will gain more velocity as a result of recoil.

Explanation:

Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''

Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.

Answer:

[tex]T=3*10^-3 N/m[/tex]

Explanation:

From the question we are told that:

Radius :

[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]

Work [tex]W=1.5 * 10^{-4}[/tex]

Generally the equation for Work done  is mathematically given by

[tex]W=TdA[/tex]

Where

[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]

[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]

[tex]dA=0.050m^2[/tex]

Therefore

[tex]W=TdA[/tex]

[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]

[tex]T=3*10^-3 N/m[/tex]

Answer:

(a) 14.14 s

(b) 4.24 s

Explanation:

maximum acceleration, a =  1 m/s^2

maximum speed, v = 5 m/s

initial speed, u = 0 m/s

(a) distance, s = 100 m

Let the time is t.

Use second equation of motion

[tex]s = u t 0.5 at^2\\\\100 = 0 + 0.5 \times 1 \times t^2\\\\t = 14.14 s[/tex]

(b) distance, s = 9 m

Let the time is t'.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\9= 0 + 0.5 \times 1 \times t'^2\\\\t' = 4.24 s[/tex]

Answer:

hgff

Explanation:

Answer:

The charge moves to equilibrium.

E.e = B.e.V

E is electric field force.

e is the charge.

B is magnetic field force.

V is acceleration voltage.

Answer:

First remember that the distance between two points (a, b) and (c, d) is given by the equation:

[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]

Now let's define the position of the radar as:

(0mi, 0mi)

Then we can write the position of the plane as:

(480mi/h*t, 1mi)

where t is time in hours.

Then we can write the distance equation as:

[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]

Now we want to get:

the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.

So first we want to find the value of t such that:

d(3) = 3mi

We will look at the positive value of t, because at this point the plane is increasing its distance to the station.

[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]

The rate of change when the plane is 3 mi away from the station is:

d'(0.0059h)

remember that:

d'(t) = dd(t)/dt

We can write:

d(t) = h( g(t) )

such that:

h(x) = √x

g(t) = (480mi/h*t)^2 + (1mi)^2

then:

d'(t) = h'(g(t))*g'(t)

This is:

[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]

The rate of change at t = 0.0059h is then:

[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]

Answer:

True

Explanation:

This is because of the point where the forces are applied by our muscles and

the angle they have about the bones. Take for example the  diagram I uploaded.

If we do a free body diagram and a sum of torques, we would get that:

[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]

In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:

[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]

If we solve the equation for the force of the muscle we would get that:

[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]

Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.

Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:

[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]

which yields:

[tex]F_{muscle}=4.04 F_{hand}[/tex]

so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

If you want to learn more, you can read:

https://brainly.com/question/6692050

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

Answer:

Answer is B (Decomposition)

Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(

Answer:

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Explanation:

a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:

[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)

Where:

[tex]W[/tex] - Useful work, in joules.

[tex]E[/tex] - Food energy, in joules.

If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:

[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]

[tex]\eta = 9.082\,\%[/tex]

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:

[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]

[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]

[tex]E = 7.60\times 10^{5}\,J[/tex]

[tex]E = 181.644\,kcal[/tex]

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Answer:

log T = 3.72 to 3.77

Explanation:

Temperature range is

T = 5200 to 5900

Take the log

So,

log T = log 5200 to log 5900

log T = 3.72 to 3.77

Answer:

Explanation:

When a falcon is hovering, the force of up thrust is balanced by the weight.

When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

[tex]4.4×10^{-7}\:\text{m}[/tex]

Explanation:

The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:

[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]

or

[tex]\dfrac{hc}{\lambda} = \phi[/tex]

Solving for the wavelength [tex]\lambda[/tex],

[tex]\lambda = \dfrac{hc}{\phi}[/tex]

[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]

[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]

Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.

The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.

To find the wavelength, the given values are,

Energy (E) = 4.50×10¯¹⁹ J

The distance between two consecutive crests and troughs is called the wavelength of a wave.

Here, for the wavelength,

Energy (E) = 4.50×10¯¹⁹ J

Planck's constant (h) = 6.626×10¯³⁴ Js

Speed of light (v) = 3×10⁸ m/s

The wavelength of the light can be obtained as illustrated below:

E = hv / λ

Cross multiply λ,

E × λ = hv

Divide both sides by E,

λ = hv / E

Substituting all the values,

λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹

λ = 0.000000441733 m

λ = 441.73nm

λ - The maximum wavelength of light.

Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm

Learn more about wavelength,

https://brainly.com/question/13047641

#SPJ2

Answer:

your answer is -788231

Explanation:

1+1677=1678

1678-789909=-788231

Complete Question

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2

Answer:

[tex]h=1614m[/tex]

Explanation:

From the question we are told that:

Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]

Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]

Density [tex]\rho=1.20kg/m^2[/tex]

Generally the equation for Height climbed is mathematically given by

[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]

[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]

[tex]h=1614m[/tex]