How much energy is stored in the capacitor before the dielectric is inserted?Express your answer in joules.

Answer:

The critical angle is 34 degree.

Explanation:

Given the index refraction = 1.77

We have to find the critical angle of sapphire when the air has the index of refraction of 1.0.

Use the below formula to find the angle.

sin(θ)c = 1 / n

sin(θ) c  = 1 / 1.77

(θ) c  = Sin^-1 (0.565)

= 34.4 degree

= 34 degrees.

Thus, the critical angle is 34 degree.

"Acceleration" means the rate at which velocity is changing.

If velocity is constant, then it isn't changing.

If velocity isn't changing, then acceleration is zero.

Answer:

Weight of the woman in Newton

   [tex]x = 698.6 \ N[/tex]

 Mass of the woman in slug

  [tex]Mass = 4.86 \ slug[/tex]

 Mass of the woman in kg

   [tex]Mass = 16 \ kg[/tex]

  My weight in Newton

  [tex]W = 784 \ N[/tex]

Explanation:

From the question we are told that

   The weight of the woman in pounds is  [tex]W = 157 \ lb[/tex]

Converting to Newton

    1 N  =  0.22472 lb

    x N  =  157

=>   [tex]x = \frac{157 * 1}{0.22472}[/tex]

=>   [tex]x = 698.6 \ N[/tex]

Obtaining the mass in slug

   [tex]Mass = \frac{W}{g}[/tex]

Here  [tex]g = 32.2 ft/s^2[/tex]

So

     [tex]Mass = \frac{157 }{32.2}[/tex]

       [tex]Mass = 4.86 \ slug[/tex]

Obtaining the mass in kilogram

     [tex]Mass = \frac{W}{g}[/tex]

Here  [tex]g = 9.8 \ m/s[/tex]

So

   [tex]Mass = \frac{157 }{9.8}[/tex]

   [tex]Mass = 16 \ kg[/tex]

Generally weight is mathematically represented as

     [tex]W = m * g[/tex]

Given that my mass is   80 kg   then my weight is  

     [tex]W = 80 *9.8[/tex]

      [tex]W = 784 \ N[/tex]

Answer:

science is the knowledge gained through observations and experimentations

Characteristics

consistency

observability

predictability

testability

tentativeness

components of nature are soil,atmosphere and radiation

"Should wolves be reintroduced into national parks" is a question that cannot be answered through making measurements, therefore the correct answer is option C.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

A length, for instance, is a physical quantity. A defined, predetermined length is represented by the length unit known as the meter.

A qualitative type of question cannot be answered while making a quantitative measurement.

Thus, the question of whether wolves should be reintroduced into national parks cannot be answered by taking measurements, so option C is the appropriate response.

Learn more about the unit of measurement from here

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#SPJ2

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

[tex]2d\sin\theta=n\lambda[/tex]

For second order, n = 2

[tex]\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)[/tex]

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

Answer:

a

  [tex]F = 2.32*10^{-17} \ N[/tex]

b

[tex]n =3869 \ electrons[/tex]

Explanation:

From the question we are told that

   The  charge on each water drop is  [tex]q_1=q_2=q = - 6.19*10^{-16} \ C[/tex]

   The  distance of separation is  [tex]d = 1.22\ cm = 0.0122 \ m[/tex]  

Generally the electrostatic force between the water drops is mathematically represented as

      [tex]F = \frac{k * q_1 * q_2 }{ d^ 2}[/tex]

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

So  

      [tex]F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}[/tex]

       [tex]F = 2.32*10^{-17} \ N[/tex]

Generally the quantity of charge is mathematically represented as

        [tex]q = n * e[/tex]

Here n is the number of electron present

and  e is the charge on one electron with value  [tex]e = 1.60*10^{-19} \ C[/tex]

    So

         [tex]n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}[/tex]

         [tex]n =3869 \ electrons[/tex]

Answer:

(c) is accelerating upward

Explanation:

When a person stand in an elevator moving upwards, he feels heavier because the elevator's floor presses harder on his feet, and the scale will show a higher reading than when the elevator is at rest.

According to Newton's second law

R = mg + ma

where;

R is the reading of the scale = apparent weight of the person

mg is the normal weight of the person

ma is the upward force acting on the person

Therefore, his apparent weight will be the greatest when the elevator is accelerating upward

(c) is accelerating upward

Answer:

1.14*10^-3m

Explanation:

We know that

Volume= mass/ density

But

Area x depth= mass/density

πr²*depth= mass/density

r²=m/πdp

So

r= √0.35/3.142*0.085*1000

r= 1.14*10^-3m

Answer:

The new period is 16 s

Explanation:

The period of a simple pendulum is given by

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]

Where [tex]T[/tex] is the period

[tex]L[/tex] is the length of the string

[tex]g[/tex] is the acceleration due to gravity

2π is constant

For the first experiment,

[tex]T =8 s[/tex]

[tex]g = gE[/tex]

Then,

[tex]8 = 2\pi \sqrt{\frac{L}{gE} } \\[/tex]

[tex]8\sqrt{gE} = 2\pi \sqrt{L}[/tex] ....... (1)

For the second experiment

[tex]T = T_{2}[/tex]

[tex]g = \frac{1}{4} gE[/tex]

Hence,

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex] becomes

[tex]T_{2} =2\pi \sqrt{\frac{L}{\frac{1}{4}gE } }[/tex]

Then,

[tex]T_{2} \sqrt{\frac{1}{4}gE } = 2\pi\sqrt{L}[/tex] ....... (2)

Since 2π is constant and

The same pendulum is used for the second experiment, then [tex]L[/tex] is also constant.

∴ [tex]2\pi \sqrt{L}[/tex] is constant for both experiment.

Hence, we can equate equations (1) and (2) such that

[tex]8\sqrt{gE} = T_{2} \sqrt{\frac{1}{4}gE }[/tex]

Then,

[tex]\frac{8\sqrt{gE} }{\sqrt{\frac{1}{4}gE } } = T_{2} \\\frac{8\sqrt{gE} }{\frac{1}{2}\sqrt{gE} } } } = T_{2} \\ 16 = T_{2} \\T_{2} = 16 s[/tex]

Hence, the new period is 16 s

A simple pendulum swings for 8 seconds. The new period is determined by repeating the pendulum experiment on a plant with g = 1/4 gE.

Move to a position where the gravitational acceleration is greater.

- Reduce the pendulum's length.

Explanation: The frequency of a simple pendulum is given by where L is the length of the pendulum g is the gravitational acceleration

From the formula, we notice that - As the value of g increases, the frequency of the pendulum increases as well

- As the value of L increases, the frequency of the pendulum decreases

So, in order to increase the frequency of the pendulum, we can: - Move to a location where the acceleration due to gravity, g, is larger
To learm more about pendulum's visit: https://brainly.com/question/12495919

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

Explanation:

Answer:

entropy

Explanation:

GPG is an acronym for GNU Privacy Guard, which is considered as free software, is a command-line tool for generating cryptographically secure keys.

These cryptographically secure keys can be created by using entropy, through reliable entropy source, like /dev/random, which in turn derives its entropy from physical devices connected to the machine, such as keyboards, mice, and disks.

Hence, the right answer is ENTROPY

Answer:

__________________

I think the answer is A.

__________________

Answer:

The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

[tex]E_{i}=E_{f}[/tex]

[tex]\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2[/tex]

[tex]\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})[/tex]

[tex]0.00075(400-v_{f}^2)=0.18 [/tex]

[tex]400-v_{f}^2=\dfrac{0.18}{0.00075}[/tex]

[tex]-v_{f}^2=240-400[/tex]

[tex]v_{f}^2=160[/tex]

[tex]v_{f}=4\sqrt{10}\ m/s[/tex]

Hence, The speed of q₂ is [tex]4\sqrt{10}\ m/s[/tex]

Answer:

A

Explanation:

Gravitational force given by

F_ps =G(m_pM_s)/r^2

G= gravitational constant

m_p=mass of the probe =125 Kg

M_s= mass of sun

r= distance between probe and Sun.

Similarly, F_pe = G(m_pM_e)/r^2.

m_p = mass of probe

M_e = mass of earth

Clearly, M_s> M_e . Therefore, F_ps> F_pe.

Hence,  A. The force is toward the sun.

Answer:

0.98°

Explanation:

First we find the refracting angle of orange in water

Which is given as

စr= sin^-1. ( 1.456 sin60°/1.33)

= 71.2°

Then that of violet in water

စv=sin^-1. ( 1.468sin60°/1.342)

= 72.3°

So angle between boths colours is the difference

71.2- 72.3= 0.98°

Answer:

Explanation:

An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.

The change in moment is given by the expression below

Δp=m⋅(Δv)=m⋅(vf−vi) .

Given data

mass m= 0.140-kg

initial velocity vi= 120 m/s

final velocity vf= 1 m/s

substituting we have

Δp=m⋅(Δv)=0.14⋅(1−1.2)

Δp=m⋅(Δv)=0.14⋅(-0.2)

Δp= 0.028 kg m/s

The change in momentum was found to be Δp= 0.028 kg m/s

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum  magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

=  0

position at t = 1 s

= 10 +0

= 10  m

So range of position is 8.5 m to 10 m .

We want to find the range of possible positions for the cart at t = 1s.

The range is:

14.5m ≤ p ≤ 16m

First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:

a(t) = a

To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.

v(t) = a*t + 3m/s

To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.

p(t) = (a/2)*t^2 + (3m/s)*t + 10m

Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.

The minimum acceleration is a = 3m/s^2, so we have the minimum position:

p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m

The maximum acceleration is 6m/s^2, so the maximum position is:

P(1s) =  (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m

So the range of the position at t = 1s is:

14.5m ≤ p ≤ 16m

If you want to learn more about motion equations, you can read:

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(1) -2.0025m/s.

(2) It moves west.

(3) 3.3375m/s.

(4) It moves east.

Assuming the collision occurred in an isolated system. This means that total momentum of the system of pucks is conserved. Since no external forces are acting on these pucks, the momentum of the pucks before collision is equal to the momentum of the pucks after collision. i.e

(p₁)₀ + (p₂)₀ = (p₁)₁ + (p₂)₁              -------------(i)

Where;

(p₁)₀ = momentum of the 0.45kg puck before collision

(p₂)₀ = momentum of the 0.990kg puck before collision

(p₁)₁ = momentum of the 0.45kg puck after collision

(p₂)₁ = momentum of the 0.990kg puck after collision

But;

(p₁)₀ = m₁ u₁  

[m₁ = mass of the 0.45kg, u₁ = speed of the 0.45kg before collision]

(p₂)₀ = m₂u₂

[m₂ = mass of the 0.990kg, u₂ = speed of the 0.990kg before collision]

(p₁)₁ = m₁v₁

[m₁ = mass of the 0.45kg, v₁ = speed of the 0.45kg after collision]

(p₂)₁ = m₂v₂

[m₂ = mass of the 0.990kg, v₂ = speed of the 0.990kg after collision]

Equation (i) then becomes;

m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂    ----------------(ii)

From the question:

m₁ = 0.450kg

u₁ = +5.34m/s   [Taking east direction as positive]

m₂ = 0.990kg

u₂ = 0m/s    [since the second puck is initially at rest]

Substitute these values into equation (ii)

(0.450 x 5.34) + (0.990 x 0) = 0.45 v₁ + 0.990 v₂

2.403 + 0 = 0.45 v₁ + 0.990 v₂

2.403 = 0.45 v₁ + 0.990 v₂           ------------------(iii)

Also, since the collision is perfectly elastic, it means that the kinetic energy is conserved. i.e the total kinetic energy before collision is equal to the total kinetic energy after collision.

=> [tex]\frac{1}{2}[/tex]m₁ u²₁ + [tex]\frac{1}{2}[/tex]m₂u²₂ = [tex]\frac{1}{2}[/tex]m₁v²₁ + [tex]\frac{1}{2}[/tex]m₂v²₂

Substitute the necessary values into the above equation:

[[tex]\frac{1}{2}[/tex] x 0.45 x 5.34²] + [0] = [[tex]\frac{1}{2}[/tex] x 0.45 x v²₁] + [[tex]\frac{1}{2}[/tex] x 0.990 x v²₂]

[6.41601] = [0.225 x v²₁] + [0.495 x v²₂]  ------------------(iv)

Now let's solve equations (iii) and (iv) simultaneously

2.403 = 0.45 v₁ + 0.990 v₂

6.41601 = 0.225 x v²₁ + 0.495 x v²₂

let

v₁ = x

v₂ = y

2.403 = 0.45 x + 0.990 y                  ------------(5)

6.41601 = 0.225 x² + 0.495 y²         -------------(6)

From equation (5), make x subject of the formula

2.403 = 0.45x + 0.990y

0.45x = 2.403 - 0.990y         [divide through by 0.45]

x = 5.34 - 2.2y    ----------------(m)

Substitute x into equation (6)

6.41601 = 0.225 (5.34 - 2.2y)² + 0.495 y²     [expand bracket]

6.41601 = 0.225 [28.5156 - 23.496y + 4.84y²] + 0.495 y²   [remove bracket]

6.41601 = 6.41601 - 5.2866y + 1.089y² + 0.495 y²

1.584y² - 5.2866y = 0

y(1.584y - 5.2866) = 0

y = 0       or       1.584y - 5.2866 = 0

y = 0       or       1.584y = 5.2866

y = 0       or       y = 3.3375

Since y = v₂ cannot be zero because the puck will definitely move after collision, the second value of y = 3.3375 is considered.

Substitute this value into equation (m)

x = 5.34 - 2.2y

x = 5.34 - 2.2(3.3375)

x = 5.34 - 7.3425

x = -2.0025

Therefore,

x = v₁ = -2.0025m/s

y = v₂ = 3.3375m/s

(1) From the analyses above, the speed of the 0.450kg puck after collision is -2.0025m/s.

(2) Since the speed is negative, it shows that the 0.45kg puck moves opposite the direction at which it was moving before collision. It moves west.

(3) The speed of the 0.990kg puck after collision is 3.3375m/s.

(4) Since the speed is positive, it shows that the 0.990kg puck moves east. Remember that east has been taking as the positive direction.

Answer:

A. a meteor traveling unhindered through space

Explanation:

Answer:

5.2 centimicro meter or 5.2 cμm

Explanation:

Prefixes allow scientists report numbers logically depending on the order of the magnitude of the numbers. For example to represent 4 x 10⁻³m, it is more handy to say : 4 millimetre, where milli- is the prefix for 10⁻³.

Common prefixes are:

deci = 10⁻¹                     deca = 10¹                    

centi = 10⁻²                   hecto = 10²

milli  = 10⁻³                    kilo  = 10³                      

micro = 10⁻⁶                  mega = 10⁶

nano = 10⁻⁹                   giga = 10⁹

pico = 10⁻¹²                   terra = 10¹²

femto = 10⁻¹⁵                peta = 10¹⁵

Now, let's represent the given diameter: 5.2 x 10⁻⁸ m

Since the 10⁻⁸ prefix is not common (or maybe doesn't even exist), the way around this is to break it into bits such as:

10⁻⁸ = 10⁻² x 10⁻⁶           [10⁻² = centi,  10⁻⁶ = micro ]

10⁻⁸ = centimicro

Therefore, 5.2 x 10⁻⁸ m becomes 5.2 centimicro meter.

The symbol is 5.2 cμm

Where c = centi, μ = micro.

Answer:

Average speed = 2.22m/s

Velocity = 0.89 m/s west

Explanation:

(i) Since speed is a scalar quantity, the direction of movement by the person is irrelevant. Therefore,

Average speed = total distance traveled / time taken

Where;

Total distance = 70m + 30m = 100m

Time taken = 45 seconds

Average speed = 100 / 45 = 2.22m/s

(ii) However, velocity is a vector quantity and so the direction of movement by the person is of utmost importance. Therefore,

Velocity = displacement / time taken

Where;

displacement = -70m + 30m    [West direction is taken as negative and east is taken as positive]

displacement = -40m     [The negative sign just means that the net displacement is in the direction of the West].

Thus, displacement can be written as 40m west.

Therefore,

Velocity = 40 / 45 = 0.89m/s west

Answer:

m = 312.5

Explanation:

Given that,

The focal length of the objecive lens, [tex]f_o=1\ cm[/tex]

The focal length of eye piece, [tex]f_e=2\ cm[/tex]

length of the tube, L = 25 cm

We need to find the magnitude of the overall magnification of the microscope. It is given by the formula as follows :

[tex]m=\dfrac{L}{f_o}\times \dfrac{D}{f_e}[/tex]

D = 25 cm

So,

[tex]m=\dfrac{25}{1}\times \dfrac{25}{2}\\\\m=312.5[/tex]

So, the overall magnification of the microscope is 312.5.

Explanation:

Given:

v₀ = 0 m/s

a = 2.50 m/s²

t = 4 s

Find: v

v = at + v₀

v = (2.50 m/s²) (4 s) + 0 m/s

v = 10 m/s

The speed of the ball after 4s is 10 m/s.

There's not enough information to find its velocity. (We would need to know what direction it's moving.)

Answer:

in scientific notation 26500 is 2.65×10⁴

Answer:

2.56*10^4

26500

(We have to move the decimal 4 places)

2.6500*10^4

So the answer is 2.6500*10^4 or 2.56*10^4

Hope this helps! :)

Answer:

40 Ω

Explanation:

The following data were obtained from the question:

Potential difference (V) = 20 V

Current (I) = 0.50 A

Resistance (R) =?

From ohm's law:

V = IR

Where:

V is the potential difference.

I is the current.

R is the resistance.

With the above formula, we can obtain the resistance of the resistor as follow:

Potential difference (V) = 20 V

Current (I) = 0.50 A

Resistance (R) =?

V = IR

20 = 0.5 × R

Divide both side by 0.5

R = 20/0.5

R = 40 Ω

Therefore, the resistance of the resistor is 40 Ω

Answer:

175 N/m

Explanation:

Given:

Force = F=  14.0 N

Distance = x = 8.00 cm = 0.08 m

To find:

spring constant

Solution:

spring constant is calculated by using Hooke's law:

k =  F/x

Putting the values in above formula:

k = 14.0 / 0.08

k = 175 N/m

Answer:

5.68*10^-5 °C

Explanation:

ADC is an acronym that means analog to digital conversion.

The numbers of bit in any ADC is the level of Voltage it possess.

The question states 32 bits, this means that it's total voltage levels is 2^32

Also, the question gives us a range of 10 V for the ADC. Now we can evaluate the resolution of the voltage as:

Δv = 10 / (2^32)

Δv = 10 / 4.29^9

Δv = 2.33*10^-9

The thermocouple sensitivity is 41 micro volts/°C

Thermocouple sensitivity = Δv/ΔT, where ΔT is the temperature change

ΔT = Δv/thermocouple sensitivity

ΔT = 2.33*10^-9 / 41*10^-6

ΔT = 5.68*10^-5 °C

Thus, the smallest temperature change is 5.68*10^-5 °C

Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d  / M  

u² = 2 x E₁ e d  / M

Similarly for electrons

u² = 2 x E₂ e d  / m

Hence

2 x E₁ e d  / M =  2 x E₂ e d  / m

E₁ / E₂ = M / m