Two slits are separated by 0.370 mm. A beam of 545-nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range−26.0° ≤ θ ≤ 26.0°.

Answer:

Explanation:

Total mass m = 18 kg .

Spring are parallel to each other so total spring constant

= 4 x 24 = 96 N/cm = 9600 N/m

Time period of vibration

[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]

Putting the given  values

[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]

= .27 s .

Answer:

20.07ohms

Explanation:

Impedance is defined as the opposition to the flow of current through the elements of the circuit.

Impedance for R-L AC circuit is expressed as Z = √R²+XL²

R is the resistance

XL is the inductive reactance.

Given resistance of the air condition = 20 ohms

Inductive reactance XL = 1.68 ohms

Z = √20²+1.68²

Z = √400+2.8224

Z = √402.8224

Z = 20.07 ohms

Hence the impedance of the air conditioner is 20.07ohms

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]

The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

[tex]a_x = \frac{d^2 x }{dt^2}[/tex]

[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]

Similarly,

[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]

[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]

[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

Answer:

B.solar panels generate electricity that keeps the satellites running.

Explanation:

Solar panels are a renewable resource because they take energy from the sun.

Answer:

5.81 X 10^3 Ns

Explanation:

Given that

F = At² and F at t = 1.25 s is 781.25 N ?

A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²

d(Impulse) = Fdt

Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s

Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s

Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns

5.81 X 10^3 N.s

Answer:

E = 1.24MeV

Explanation:

The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.

Its energy can be determined by;

E = hf

  = (hc) ÷ λ

where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.

Therefore,

E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]

  = 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]

E = 1.9878 × [tex]10^{-13}[/tex] J

But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;

E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]

  = 1242375 eV

∴ E = 1.24MeV

The energy of the photon is 1.24MeV.

Answer:

5.33333... seconds

Explanation:

800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....

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 .

The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]

The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.

According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.

Mathematically, the net work done (W) on an object or physical body is given by the formula:

[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]

Where:

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Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

Answer:

gravitational potential energy at point A.

A) The gravitational potential energy at point A is; 705600 J

B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s

A) Formula for gravitational potential energy is;

PE = mgh

At point A;

mass; m = 900 kg

height; h = 80 m

Thus;

PE = 900 × 9.8 × 80

PE = 705600 J

B) Kinetic energy of the roller coaster at point C is given as;

KE = PE - W

We are given Workdone; W = 264870 J

Thus;

KE = 705600 - 264870

KE = 440730 J

Thus, velocity at point C is gotten from the formula of kinetic energy;

KE = ½mv²

v = √(2KE/m)

v = √(2 × 440730/900)

v = 31.295 m/s

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Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

[tex] M = log(\frac{I}{S}) [/tex]

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

Answer:

Explanation:

Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.

Answer:

a)  A = 449526  J,  b) 449526 J

Explanation:

In this exercise they do not ask for the work of different elements.

Note that as the box rises at constant speed, the sum of forces is chorus, therefore

           T-W = 0

           T = W

           T = m g

           T = 1,390 9.8

           T = 13622 N

Now that we have the strength we can use the definition of work

           W = F .d

            W = f d cos tea

a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel

              A = A  x

              A = 13622  33

               A = 449526  J

b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180

               W = T x cos 180

               W = - 13622 33

               W = - 449526 J

Answer:

The frequency will be reduced by a factor of √2/2

Explanation:

Pls see attached file

The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

Let the initial mass of block be m.

And new mass is, 4m.

The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,

[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

Here,

k is the spring constant.

If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,

[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]

Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

Learn more about the frequency of oscillation here:

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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:

Let the acceleration be a of the system

T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass

for motion of 3.85 kg , applying newton's law

3.85g - T₁ = 3.85 a

for motion of 2.6 kg

T₂ - 2.6g = 2.6 a

T₂ - T₁ + 1.25 g = 6.45 a

T₁ - T₂ = 1.25 g -  6.45 a

for motion of pulley

(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration

(T₁ - T₂ ) x R = 1 /2  m R² x a / R

(T₁ - T₂ )  =   m  x a / 2 = .79 x a / 2 = . 395 a

1.25 g -  6.45 a = .395 a

1.25 g = 6.845 a

a = 1.79 m /s²

B )

When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s

Average deceleration = .2 / 6.4 = .03125 m /s²

Total deceleration created by frictional torque = 1.79 + .03125

= 1.82125 m /s²

If R be the average frictional torque  acting on the pulley

angular deceleration of pulley = a / R

= 1.82125 / .045

= 40.47 rad /s²

Now  R = I x 40.47 , I is moment of inertia of pulley

= 1 /2 x .79 x .045² x 40.47

= .0323 N.m

Torque created = .0323 Nm

The acceleration of the two masses hanging from ends of the pulley is 31 m/s².

The average frictional torque acting on the pulley is 0.55 Nm.

The given parameters;

The acceleration of the two masses is determined by taking net torque acting on the pulley;

[tex]\tau _{net} = I \alpha[/tex]

[tex]T_1R - T_2R = I \alpha\\\\[/tex]

where;

[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]

Substitute the given parameters, to solve for the acceleration of the masses;

[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]

The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.

[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]

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one of the answers that i found was   5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.

Answer:

Your answer is( D) - Arago

Answer:

0.9m/s²

Explanation:

See attached files

The number of maxima of the standing wave pattern is two.

At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.

learn more about the waves here: https://brainly.com/question/3004869?referrer=searchResults

Answer:

Car 3

Explanation:

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

Answer:

m = 300668.9 kg

L₀ = 12.47 m

Explanation:

The relativistic mass of the space vehicle is given by the following formula:

[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]

where,

m = relativistic mass = ?

m₀ = rest mass = 150000 kg

v = relative speed = 2.6 x 10⁸ m/s

c = speed of light = 3 x 10⁸ m/s

Therefore

[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]

m = 300668.9 kg

Now, for rest length of vehicle:

L = L₀√(1 - v²/c²)

where,

L = Relative Length of Vehicle = 25 m

L₀ = Rest Length of Vehicle = ?

Therefore,

25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]

L₀ = (25 m)(0.499)

L₀ = 12.47 m

Answer:

Mb²/2

Explanation:

Pls see attached file

Answer:

D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity

Explanation:

Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.

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