When you shine a beam of light, which is composed of just two different colors, red and green, onto a diffraction grating which color gets diffracted more

Answer:

C. To draw on an earlier experience.

Complete question:

A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.

a. 0.06 T

b. 0.08 T

c. 0.07 T

d. 0.05 T

Answer:

The magnitude of the magnetic field is 0.07 T.

Explanation:

Given;

magnitude of the charge, q = 15C

velocity of the charge, v = 6.2 x 10³ m/s

angle between the charge and the magnetic field, θ = 48°

the force on the particle, F = 4838 N

The magnitude of the magnetic field can be calculated by applying Lorentz force formula;

F = qvBsinθ

where;

B is the magnitude of the magnetic field

B = F / vqsinθ

B = (4838) / (6.2 x 10³ x 15 x sin48)

B = 0.07 T

Therefore, the magnitude of the magnetic field is 0.07 T.

Answer:   2

Explanation:

1/2=1/1 +1/x

x=-2

magnification= 2/1

magnification=2

The ratio of foort dustance to load distance in a simple machine is called mechanical advantage or MA.

MA= Effort Distance / Load Distance

The question is incomplete, the complete question is;

how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3  m 2  and a plate separation of 2.00 mm.

The capacitance of the capacitor is the quantity of charge stored by the capacitor.

Given that;

C1= εo k * A/d

εo = permittivity of free space

C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3  m 2/2 * 10^-3

= 2.21 * 10^-11 F

C2 = 8.85 x 10-12 * 3 * 5.00×10 −3  m 2/2 * 10^-3

= 6.63 * 10^-11 F

q1 = C1V1 = 2.21 * 10^-11 C * 12 V

= 2.65 * 10^-10 C

q2 = C2V2 = 6.63 * 10^-11 F * 12 V

= 7.96 * 10^-10 C

qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C

qtotal = 1.061 * 10^-9C

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Answer:

V = (k*Q)/R

Explanation:

Total potential difference accelerates the electron to being very far away from Q is;

V = (k*Q)/R

Where,

V is the Potential Difference in Joules per Coulomb

k is the constant

Q is the charge in Coulomb

R is Electron distance in cm or m

Example

An electron starts from rest 66.1 cm from a fixed point charge with Q = -0.120 μC. What total potential difference accelerates the electron from being very far away from Q

For k = 9.0*10^9 N*m^2/C^2

V = (k*Q)/R

V = (9.0*10^9 * -0.120*10^-6)/0.661

V = -1633.9 Volt.

The answer will change to positive because V = (k*Q)/R is negative at the outset and Zero far away.

The electron (with a negative charge) has a positive energy in the beginning and that gets converted into a positive kinetic energy "far away".

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field  = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E  x B is along k direction or z  - axis so wave is propagating along z - axis .

The direction of motion of electromagnetic wave will be +z-direction.

Electromagnetic waves are waves that consist of the electric field and magnetic field.

The electric and magnetic fields are perpendicular to each other and the wave propagates in the direction perpendicular to both the fields.

Now, the direction of wave motion can be estimated by taking the cross-product of directional unit vectors of the electric and magnetic fields.

The electric field is in the +y direction and the magnetic field is in the +x-direction.

So, the direction of the wave will be,

[tex]i\times j=k[/tex]

Therefore, the direction of motion of electromagnetic wave will be +z-direction.

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Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

The magnitude of the electric field is 1.25 N/C

But before that the following calculations need to be done.

ε = LB.v = 0.01 m × 5 T × 1 m/s

= 0.05 V

Now

ε = ∫E.ds

here ε = Eds because E is always parallel to the side of the cube

So,

= E∫ds  ∫ds

= 4L so we have 4 sides

Now

= E(4L)

= 4EL

So,4EL = 0.05 V

Now

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

hence, The magnitude of the electric field is 1.25 N/C

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Answer:

Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.

Explanation:

Answer:

[tex]\Large \boxed{\sf kilo}[/tex]

Explanation:

kilo is a prefix that means [tex]1000[/tex] of the base unit.

Answer:

kilo is the correct answer

Explanation:

because my exam says sooo....

Answer:infrared radiation

Explanation:

Infrared radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.

If we arrange   electromagnetic wave with decrease in wavelength, we get:

Radio waves > microwave >  Infrared >  Visible light > Ultraviolet > X-rays > Gamma radiation.

Hence,  Infrared  radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

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Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

           [tex]w = \frac{2 \pi}{T}[/tex]

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

I believe the correct answer is a

The true statement is that, at resonance, current is in phase with the driving voltage.

Resonance occurs in an RLC circuit at the point when the inductive reactance is equal to the capacitive reactance. At the point of resonance, the voltage across the inductor  and capacitor are equal.

It should noted that, at resonance, the current is in phase with the driving voltage. This is the true statement in the list.

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Answer:

It bends away from the normal

Explanation:

From Snell's law of Refraction, when a ray passes from a medium of lower Refractive index to a medium with higher Refractive index, the Refractive ray will bend towards the normal. However, when the ray passes from a medium of higher Refractive index to a medium of lower Refractive index, the Refractive ray will bend away from the normal.

Now, from the question we are told that Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2.

This means from a higher Refractive index to a lower one and from Snell's law as earlier said, the refracted ray will bend away from the normal

The refracted ray is seen to bend away from the normal.

Let us recall that an optically denser medium will have a higher refractive index. This means that the medium with a refractive index of  1.3  is the denser medium and the medium with a refractive index of  1.2 is the less dense medium.

From the statement in the question, we can boldly say that light is travelling from a denser to less dense medium given the values of the refractive index given. When light is travelling from a denser to a less dense medium, the refracted ray bends away from the normal.

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We're given

[tex]\displaystyle y_1(x) = 1 + \frac x3 + \frac{x^2}{24} + \frac{x^3}{360} + \cdots = \sum_{n=0}^\infty a_nx^n[/tex]

so let's see if we can find a closed form for the n-th term's coefficient.

Notice that

[tex]\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1\times3} \\\\ a_2 = \frac1{24} = \frac1{(1\times3) \times (2\times4)} \\\\ a_3 = \frac1{360} = \frac1{(1\times3) \times (2\times4) \times (3\times5)}[/tex]

If the pattern continues, the next few terms are likely

[tex]\displaystyle a_4 = \frac1{8640} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6)} \\\\ a_5 = \frac1{302400} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7) \times (6\times8)}[/tex]

which leads up to the n-th term,

[tex]\displaystyle a_n = \frac1{(1\times3) \times (2\times4) \times \cdots \times (n\times(n+2))} = \frac2{n!(n+2)!}[/tex]

where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.

So we have

[tex]\displaystyle y_1(x) = \sum_{n=0}^\infty \frac2{n!(n+2)!} x^n[/tex]

Now we use reduction of order to find a linearly independent solution of the form [tex]y_2(x) = v(x)y_1(x)[/tex], with derivatives

[tex]\displaystyle \frac{\mathrm dy_2}{\mathrm dx} = v(x) \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm dv}{\mathrm dx} \\\\ \frac{\mathrm d^2y_2}{\mathrm dx^2} = v(x) \frac{\mathrm d^2y_1}{\mathrm dx} + 2 \frac{\mathrm dv}{\mathrm dx} \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm d^2v}{\mathrm dx^2}[/tex]

Substitute [tex]y_2[/tex] and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :

[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} + \left(x\frac{\mathrm d^2y_1}{\mathrm dx^2}+3\frac{\mathrm dy_1}{\mathrm dx}-y_1\right)v = 0[/tex]

but since we know [tex]y_1(x)[/tex] satisfies the original DE, the last term vanishes and we're left with

[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} = 0[/tex]

Reduce the order by substituting [tex]w(x)=\dfrac{\mathrm dv}{\mathrm dx}[/tex] to get yet another DE in w(x) :

[tex]\displaystyle x y_1 \frac{\mathrm dw}{\mathrm dx} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)w = 0[/tex]

This equation is separable:

[tex]\displaystyle \frac{\mathrm dw}w = - \frac{2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1}{xy_1}\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}\frac{\mathrm dy_1}{\mathrm dx} + \frac3x\right)\,\mathrm dx[/tex]

From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for [tex]y_2(x)[/tex] by multiplying [tex]y_1(x)[/tex] by v(x). Using the fundamental theorem of calculus, you would find

[tex]\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = \frac{C_1}{x^3} \exp\left(-2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)} \,\mathrm d\xi\right)\right) \\\\ v = C_1 \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2[/tex]

so that you end up with

[tex]\displaystyle y_2(x) = C_1 y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2y_1(x)[/tex]

But the second term is already accounted for by [tex]y_1(x)[/tex] itself, so the second solution is

[tex]\displaystyle y_2(x) = \boxed{y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega}[/tex]

You could go the extra mile and try to find a power series expression for this solution, but that's a lot of work for little payoff IMO.

Answer:

A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts. Stress and strain are both tensor quantities. ... A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts.

Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.

Explanation:

Hope dis help

Answer:

a) 3.7 x 10^-3 s

b) 230.41 rad/s

Explanation:

The angular speed N = 2200 rpm (revolution per minute)

==> 2200/60 revolutions per sec = 36.67 rps

The total angle turned in one second = 36.67 x 360° = 13201.2°

if it takes 1 sec to revolve 13201.2°

then it will take t sec to rotate 49.0°

time t = 49/13201.2 = 3.7 x 10^-3 s

conversion to rad/s = 2πN/60 = (2 x 3.142 x 2200)/60 = 230.41 rad/s

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

(C) [tex]^{208}_{84}\text{Po}[/tex]

Explanation:

[tex]^{212}_{86}\text{Rn} \rightarrow \:^4_2\text{He} + \:^{208}_{84}\text{Po}[/tex]

Answer:

0.61 m

Explanation:

The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.

using the relationship

c = fλ

where

c is the speed of light in vacuum = 3 x 10^8 m/s

f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz

λ is the wavelength = ?

λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m

answer to your question is 0.6m

Answer:

Real images are formed where rays of light actually converge, whereas virtual images occur with they are perceived to converge. Real images can be produced by passing light through converging lenses or with a concave mirror of some sort.

Explanation:

hope that helped a little :)

Real images are formed when the object is located beyond the focal length from the lens. A virtual image is formed if the object is place between focal length and pole of the converging lens.

The converging lens which forms the real images by the intersection of light rays.

Real images are produced by light passing through converging lens or a concave mirror.

The real meeting of the rays coming from the focal point of the rays after reflection. Real images are formed where rays of light actually converge, whereas virtual images occur with they are appear to converge. Real images are formed when the object is located beyond the focal length from the lens. A virtual image is formed if the object is place between focal length and pole of the converging lens.

Thus,  a convex lens produce real image from virtual object.

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

[tex]F_{pull}=m_{T}.a[/tex]

[tex]m_{T}=\frac{F_{pull}}{a}[/tex]

[tex]m_{T}=\frac{3615}{1.516}[/tex]

[tex]m_{T}=2384.5[/tex]

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For [tex]m_{1}[/tex]:

The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

[tex]m_{1} = \frac{F_{12}}{a}[/tex]

[tex]m_{1} = \frac{1387}{1.516}[/tex]

[tex]m_{1}[/tex] = 915kg

For [tex]m_{2}[/tex]:

There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]

[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]

[tex]m_{2}[/tex] = 605kg

For [tex]m_{3}[/tex]:

[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]

[tex]m_{3} = 2384.5-1520.0[/tex]

[tex]m_{3}[/tex] = 864.5kg

Answer:

1.94cm³/s

Explanation:

1L = 1000cm³

Ihr = 3600s

So

Using

Average flow rate

Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L

= 1.94cm³/s

Answer:

In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4

Granite: 2.70 × 10 32.70 × 10 3

Lead: 1.13 × 10 41.13 × 10 4

Iron: 7.86 × 10 37.86 × 10 3

Oak: 7.10 × 10 27.10 × 10 2

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

Answer:

2.98 kN, 69.1° below the horizontal

Explanation:

Sum of forces in the x direction:

Fₓ = 3 kN cos 30° − 2 kN cos 40°

Fₓ ≈ 1.07 kN

Sum of forces in the y direction:

Fᵧ = -3 kN sin 30° − 2 kN sin 40°

Fᵧ = -2.79 kN

Magnitude of the resultant force is:

F² = Fₓ² + Fᵧ²

F = 2.98 kN

The direction of the resultant force is:

tan θ = Fᵧ / Fₓ

θ = -69.1°