According to the conservation of mechanical energy, the kinetic energy before plus the kinetic energy after equals the
potential energy before plus the potential energy after.
True
False

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

Answer:

Given that

capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴

displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m

( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2

( B ) . displacement current density j

D = ε( dE / dt )the rate at which the electric field between the plates is changing(

dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer

Answer:

d) Microwave ovens emit in the same frequency band as some wireless Internet devices.

Explanation:

Microwave are radio waves of short wavelength, from about 10 centimetres to one millimetre, in the Super High Frequency and the Extremely High Frequency bands. Microwaves can penetrate into materials and deposit their energy below the surface which is why is is used in microwave heating found in microwave oven. Transmission of data sometimes involves the use of microwaves to send and receive information over a long distance. Microwaves are the mainly used in radar, used for satellite communication, and wireless networking technologies such as Wi-Fi.

Yeah

All they are all correct

Explanation:

Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]

[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]

[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]

The work function [tex]\phi[/tex] is then

[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]

[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]

The work function of the surface is equal to 3.98 × 10⁻¹⁹J.

The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.

The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.

The expression between wavelength (λ), frequency, and speed of light (c) is:

c = νλ

Given, the wavelength of the light, ν = 500 nm

The frequency of the light can determine from the above-mentioned relationship:

ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz

The work function = h ν =  6 × 10¹⁴ × 6.626 × 10⁻³⁴

φ = 3.98 × 10⁻¹⁹J

Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.

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Answer:

Around the center of the mirror

Answer:

Velocity of wave (V) = 5.2 × 10² m/s

Explanation:

Given:

Per unit length mass (U) = 4.80 × 10⁻³ kg/m

Tension (T)= 1,300 N

Find:

Velocity of wave (V)

Computation:

Velocity of wave (V) = √T / U

Velocity of wave (V) = √1300 / 4.80 × 10⁻³

Velocity of wave (V) = √ 270.84 × 10³

Velocity of wave (V) = 5.2 × 10² m/s

Answer:

The probability that Susan turns 85 years old is 45,436/97,825= 0.4644.

Explanation:

Edmentum

Answer:

45,436/97,825= 0.4644.

Explanation:

Let [tex]R_1[/tex] and [tex]R_2[/tex] be the the resistances of the resistors. We are given that

[tex]R_1 + R_2 = 690\:Ω\:\:\:\:\:\:\:(1)[/tex]

and

[tex]\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{118\:Ω}\:\:\:\:\:(2)[/tex]

From Eqn(1), we can write

[tex]R_2 = 690\:Ω - R_1[/tex]

and then plug this into Eqn(2):

[tex]\dfrac{1}{R_1} + \dfrac{1}{690\:Ω - R_1} = \dfrac{1}{118\:Ω}[/tex]

or

[tex]\dfrac{690\:Ω}{(690\:Ω)R_1 - R_1^2}= \dfrac{1}{118\:Ω}[/tex]

[tex]\Rightarrow R_1^2 - (690\:Ω)R_1 + (690\:Ω)(118\:Ω)= 0[/tex]

or

[tex]R_1^2 - 690R_1 + 81420 = 0[/tex]

Using the quadratic formula, we find that the above equation has two roots:

[tex]R_1 = 151.1\:Ω,\:\:538.9\:Ω[/tex]

This means that if you choose one root value for [tex]R_1[/tex], the other root will be the value for [tex]R_2[/tex].

Answer:

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = [tex]\infty[/tex] (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

focus point of the light that is incident and parallel to the central axis (v)

Substituting this values into the given formula we will have;

[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]

Cross multiply

[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]

Hence  Light incident parallel to the central axis is focused at a point 36cm from the surface

Answer:

The question is not complete, here is the other part.

Assume that n = 1.36.

Express your answer to three significant figures and include the appropriate units.

λ = 707.2nm

Explanation:

n = 1.36

t = 130

2 n t= (m+1/2) λ

To solve for λ.

λ = 2 n t ÷ m+1/2

λ = 2 × 1.36 × 130 ÷ m +1/2

λ =

If m= 0

λ= 353.6 ÷ 0+ 1/2

λ = 353.6 × 2

= 707.2nm

Answer:

a) 91 m/s

b) 111 m/s

Explanation:

v = u + at

v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s

v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

The star is 4.62 x 10¹⁶ miles away from Earth.

The speed of light is 186,000 miles per second. It takes light from a particular star approximately 9 years to reach Earth. There are 365 days in 1 year, so it takes 9 x 365 = 3285 days for light from the star to reach Earth.

The distance between the star and Earth is 3285 x 186,000 = 608,810,000 miles. In scientific notation, this is 4.62 x 10¹⁶ miles.

Here is the calculation:

distance = speed * time

distance = 186,000 miles/second * 3285 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute

distance = 608,810,000 miles

distance = 4.62 x 10¹⁶ miles

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Answer:

As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.

Explanation:

In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)

Answer:

Here are notes I took on semiconductor conductivity :

________________________________________________________
-A p-type semiconductor is made of a material in which electrical conduction is due to the movement of a positive charge.

-Examples of p-type dopants - boron, aluminum, gallium, indium, and thallium

Explanation:

In this case, indium only correct option being a dopant of a P-type semiconductor. Other options are N-type dopants.

Hopefully its correct !! <3

Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]

Answer:

c. 1.11 m/s down

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Assuming the balloon and projectile are originally at rest:

(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)

0 kg m/s = (90 kg) v + 100 kg m/s

v = -1.11 m/s

Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

Explanation:

According to Thermal Expansion of solids:

[tex]dl = \alpha \times l \times dt[/tex]

[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]

[tex]dl = 7.2 \times {10}^{ - 3} [/tex]

Answer:

Explanation:

The work done by brake friction changes the kinetic energy,

ASSUMING the road is level and other friction is ignored.

W = KEf - KEi = ½mvf² - ½mvi² = ½m(vi² - vf²)

W = ½(1800)(30² + 10²) = 900(900 - 100) = 720000 J = 720 KJ

Answer:

A.5.7 x 10^-12m/s²

B. 32.5V/m

C. Pls. See attached file for explanations

Explanation:

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

= -4.14k - 0.0484k

= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

Answer:

 f = - 20 cm

Explanation:

This exercise asks us for the focal length, which for a lens in air is

                  1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)

where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.

In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression

                 - 1/40 = (n-1) (1 / R₁ -1 / R₂)

                (1 / R₁ - 1 / R₂) = - 1/40 (n-1)

let's calculate

               (1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)

               (1 / R₁ -1 / R₂) = -1/20

 Now we change the construction material for one with refractive index

n = 2, keeping the radii,

              1 / f = (n-1) (1 / R₁-1 / R₂)

              1 / f = (n-1) (-1/20)

let's calculate

             1 / f = (2.00-1) (-1/20)

              1 / f = -1/20

              f = - 20 cm

Answer:

λ = 6.61 x 10⁻⁷ m = 661 nm

Explanation:

From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:

Δx = λL/d

where,

Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m

L = Distance between slits and screen = 2.12 m

d = slit separation = 0.49 mm = 0.49 x 10⁻³ m

λ = wavelength of light = ?

Therefore,

2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)

(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ

λ = 6.61 x 10⁻⁷ m = 661 nm

Answer:

An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected

The question is not complete so i have attached it.

Answer:

The light source is 2 cm from the bottom of the lamp

Explanation:

From the attached image, we can see that the parabola opens up with its vertex at the origin.

Now, the standard form of equation for a parabola is:

x² = 4ay

Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)

Thus, we have;

12² = 4a(18)

144 = 72a

a = 144/72

a = 2

Looking at the parabola again, the line of symmetry is at x = 0

Thus, axis of symmetry is at x = 0.

Thus, focus is at (0, 2)

So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm

The distance of the light source from the bottom of the lamp is 2 cm.

The given parameters;

Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;

[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]

Thus, the distance of the light source from the bottom of the lamp is 2 cm.

"Your question is not complete, it seems to be missing the following information";

the top right edge of the lamp has a coordinate of (12,18)

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Answer:

(B) a whole number of wavelengths.

Explanation:

Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by a whole number of wavelengths.

When the resultant effect of the combination of two identical waves result in their annihilation or complete cancellation of the effect of each other, destructive interference takes place. Hence to have two wave sources producing waves that have the same frequency wavelength and amplitude and which are always in phase with each other or have a constant phase difference are said to be Coherent source

Answer:

Explanation:

Let f₀ be the frequency .

energy of photons having frequency of f₀

= hf₀ where h is plank's constant

for electron to get ejected , work function should be equal to energy of photon

hf₀ = 3.55

4.1357 x 10⁻¹⁵ x f₀ = 3.55

f₀ = 8.58 x 10¹⁴ Hz .

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

(a)  the final speed of the water after contraction is 15 m/s.

(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.

The given parameters;

Let the initial area of the pipe = A₁

Apply the continuity equation to determine the final speed of the water after contraction as follows;

[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]

The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;

[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]

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Answer:

Explanation: