The intensity of the waves from a point source at a distance d from the source is I. What is the intensity at a distance 2d from the source?

Answer:

198 ms-1

Explanation:

According to the Heisenberg uncertainty principle; it is not possible to simultaneously measure the momentum and position of a particle with precision.

The uncertainty associated with each measurement is given by;

∆x∆p≥h/4π

Where;

∆x = uncertainty in the measurement of position

∆p = uncertainty in the measurement of momentum

h= Plank's constant

But ∆p= mΔv

And;

m= 1.770×10^-27 kg

∆x = 0.15 nm

Making ∆v the subject of the formula;

∆v≥h/m∆x4π

∆v≥ 6.6 ×10^-34/1.770×10^-27 × 1.5×10^-10 ×4×3.142

∆v≥198 ms-1

Answer:

Answer:

A. for every object submerged partially or completely in a fluid

Explanation:

This is following Archimedes principle which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether FULLYor PARTIALLY submerged, is equal to the weight of the fluid that the body displaces.

Explanation:

Answer:

n= speed of light in vacuum/ speed of light in the other medium.

Explanation:

If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;

n= speed of light in vacuum/ speed of light in the other medium.

Hence;

n= c/v

Where;

n= refractive index of the material

c= speed of light in vacuum

v = speed of light in another medium.

Note that the refractive index is the amount by which a transparent medium decreases the speed of light.

Answer:

valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.

I hope this helps

Answer:

The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.

Answer:

Answers a) and b) should be marked as correct.

Explanation:

Recall that the resonance in an LRC circuit occurs when the current through the circuit is at its maximum, and such takes place when the impedance  (Z) of the circuit reaches its maximum. This means that the impedance (see formula below) is at its minimum value:

[tex]Z=\sqrt{R^2+(\omega\,L-\frac{1}{\omega\,C})^2 }[/tex]

as per the impedance expression above, such happens when the term in parenthesis inside the root which contains the inductive reactance ([tex]\omega\,L[/tex]) and the capacitive reactance ([tex]1/\omega\,C[/tex]) have the same value.

Therefore, answers:

a) "The impedance of the circuit has its minimum value."

and

b) "The inductive reactance and the capacitive reactance are exactly equal to each other."

are correct answers.

(a) The impedance of the circuit has its minimum value.

(b) The inductive reactance and the capacitive reactance are exactly equal to each other

LRC series circuit consists of inductor, resistor and capacitor is series.

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{R^2 + (X_C -X_L)^2}[/tex]

where;

The impedance of the circuit is minimum when the capacitive reactance is equal to the inductive reactance.

[tex]X_C = X_L \\\\Z = \sqrt{R^2 \ + (0)^2} \\\\Z = R[/tex]

Learn more here:https://brainly.com/question/17330472

Answer:

The discharge rate is [tex]Q = 0.0192 \ m^3 /s[/tex]

Explanation:

From the question we are told that

   The  diameter is  [tex]d = 60 \ mm = 0.06 \ m[/tex]

    The  head is  [tex]h = 6 \ m[/tex]

     The  coefficient of contraction is  [tex]Cc = 0.68[/tex]

     The  coefficient of  velocity is  [tex]Cv = 0.92[/tex]

The radius is mathematically evaluated as

         [tex]r = \frac{d}{2}[/tex]

substituting values

        [tex]r = \frac{ 0.06 }{2}[/tex]

        [tex]r = 0.03 \ m[/tex]

The  area is mathematically represented as

      [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.03)^2[/tex]

      [tex]A = 0.00283 \ m^2[/tex]

 The  discharge rate is mathematically represented as

        [tex]Q = Cv *Cc * A * \sqrt{ 2 * g * h}[/tex]

substituting values

       [tex]Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}[/tex]

       [tex]Q = 0.0192 \ m^3 /s[/tex]

i

CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

Answer:

A compound

Explanation:

A compound is a substance formed when two or more elements are chemically joined

Answer:

Compound

Explanation:

A compound is a substance derived from the chemical combination of two or more elements

e.g Water ;

= [tex]H_2O\\Hydrogen\:and\:Oxygen[/tex]

Salt ;

[tex]NaCl\\Sodium\:and\: Chlorine[/tex]

Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

Answer:

I think the answer is A...I'm not sure

Explanation:

A=(6,4)

B=(-2,-1)

A-B=(6-(-2)),(4-(-1))

=(6+2),(4+1)

=(8,5)

Answer:

[tex]6-(-2)=[/tex]

[tex]6+2[/tex]

[tex]=8[/tex]

[tex]4-\left(-1\right)[/tex]

[tex]=4+1[/tex]

[tex]=5[/tex]

[tex](8,5)[/tex]

[tex]\textbf{OAmalOHopeO}[/tex]

Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

pls give brainliest

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

                      [tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]

Substitute,  [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]

                   [tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]

Hence, The new length of aluminum rod is 17.435 cm.

Learn more:

https://brainly.com/question/19495810

Answer:

s = 2.16 x 10¹¹ m

Explanation:

Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:

s = vt

where,

s = the distance between Earth and Mars = ?

v = speed of the wave = speed of light = 3 x 10⁸ m/s

t = time taken by the radio signals to reach the rover from Earth

t = (12 min)(60 s/1 min) = 720 s

Therefore,

s = (3 x 10⁸ m/s)(720 s)

s = 2.16 x 10¹¹ m

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

Answer:

b. The current in the loop always flows in a counterclockwise direction.

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.

Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.

Learn more about the magnetic field here:

https://brainly.com/question/14848188

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

79.7 days

Explanation:

Half-life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is the amount of time,

and T is the half life.

If 90% decays, then 10% is left.

A = A₀ (½)^(t / T)

0.1 A₀ = A₀ (½)^(t / 24)

0.1 = ½^(t / 24)

ln(0.1) = (t / 24) ln(0.5)

t ≈ 79.7 days

Answer:

1.24Mev

Explanation:

Using

E= hc/lambda

= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9

= 1.24Mev

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.

[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

Where:

[tex]m[/tex] - Mass of the crate, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]

[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]

The final height of the crate is:

[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]

[tex]y_{2} = 0.8\,m[/tex]

If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:

[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\mu_{k} \approx 0.548[/tex]

Then, the magnitude of the friction force is:

[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:

[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]f = 55.851\,N[/tex]

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]

Now, the final speed is cleared:

[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]

[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]

[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]

Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:

[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Answer:

tyygggghgtyhyrdfgyyyhjjillbxsrfvgygvnjj

Explanation:

cffczhxucuxoyitxohvojcdivbjv ohxc

Answer:

0.273m/s

Explanation:

first find out the meaning of 0.90×10−4m3/s

literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s

1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?

cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²

so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²

they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s

Answer:

Hey there!

The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body

Let me know if this helps :)

Answer:

there are 3 photos attached. so check

Explanation:

Answer:

Visible light intensity at the surface of the bulb (I) = 331 W/m²

Explanation:

Given:

Energy = 75 W

Radius = 6 /2 = 3 cm = 3 × 10⁻² m

Energy goes to visible light = 5% = 0.05

Find:

Visible light intensity at the surface of the bulb (I)

Computation:

Visible light intensity at the surface of the bulb (I) = P / 4A

Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²

Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)

Visible light intensity at the surface of the bulb (I) = 331 W/m²

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              [tex]D = \frac{1.22 \lambda }{d }[/tex]

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

The distance is  [tex]d = 0.00029065 \ m[/tex]

Explanation:

From the question we are told that

    The  first wavelength is  [tex]\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m[/tex]

     The  second wavelength is  [tex]\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m[/tex]

     The  difference in the  fringe pattern is  n =  1.0  

Generally the equation defining the effect of the movement of  the mirror M 2 in a Michelson interferometer is mathematically represented as

          [tex]2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n[/tex]

Here d is the mirror M 2  must be moved

substituting values

         [tex]2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0[/tex]

        [tex]d = 0.00029065 \ m[/tex]

Answer:

Explanation:

The general decay equation is given as [tex]N = N_0e^{-\lambda t} \\\\[/tex], then;

[tex]\dfrac{N}{N_0} = e^{-\lambda t} \\[/tex] where;

[tex]N/N_0[/tex] is the fraction of the radioactive substance present = 1/16

[tex]\lambda[/tex] is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

[tex]\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328[/tex]

Half life f the material is expressed as [tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]

[tex]t_{1/2} = \frac{0.693}{0.000328}[/tex]

[tex]t_{1/2} = 2,112.8 secs[/tex]

Hence, the half life of the material is approximately 2113 seconds