A sound wave of frequency 162 Hz has an intensity of 3.41 μW/m2. What is the amplitude of the air oscillations caused by this wave? (Take the speed of sound to be 343 m/s, and the density of air to be 1.21 kg/m3.)

Answer:

The new acceleration would be 9 m/s².

Explanation:

Acceleration of an object is 6 m/s²

Net force is equal to the product of mass and acceleration i.e.

F = ma

[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]

If the net force was tripled and the mass were doubled, it means,

F' = 3F

m' = 2m

Let a' is new acceleration. So,

[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]

So, the new acceleration would be 9 m/s².

Answer:

you need momentum in order to release energy. For example, if you need to push something heavy and you get a running head start, then it will be easier.

Explanation:

Answer:effective

Explanation:

Answer:

A) geometric optics, B) geometric optics , c) geometric optics ,

e) geometric optics, f)  geometric optics

Explanation:

For this exercise we must use the condition for interference and diffraction so that these phenomena are relevant the wavelength must be comparable to the gap spacing

                λ> = a

Lam when the spacing is much greater than the wavelength, the description of geometric optics is more and more exact

let's analyze each situation

a) let's find the wavelength

                v = λ f

                λ= v / f

                λ= 343/1000

                λ = 0.343 m

                 0.343 << 1m

therefore the description of the geometric optics of

b) red light passes through the pupil of the eye

red light has a wavelength of 700 num or more, the lojo pupil has a maximum of 8 me

         λ = 700 10⁻⁹ m = 7 10⁻⁷ m

         a = 8 mm 10⁻³

longitudinal is much less therefore the geometric optics is correct

c) luz azul lam = 450 nm = 450 10⁻⁹ m = 4.5 10⁻⁷ m

again the wavelength is much less than the diameter of the pupil, for which the description with the optics is generally sufficient

d) a radio A transmits up to a maximum of f = 1400 Khz = 1,400 10⁶ Hz

let's find the wavelength

          c = λf

          λ = c / f

           λ= 3 108 / 1,400 106

          λ= 2.14 102 m

in this case the wavelength is greater than the width of the gate, so the description of diffraction should be used to explain the phenomenon

e) X-rays have wavelength lam = 10-10 m

the separation of the elbow bones is of the order of a few millimeters, for local the wavelength is much less than the separation, therefore with the relations of geometric optics it is sufficient

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

Answer:

1

  [tex]R = 1.692*10^{6} \Omega[/tex]

2

  [tex]C = 2.837 *10^{-6} \ F[/tex]

Explanation:

From the question we are told that

    The voltage is  [tex]E = 110 \ V[/tex]

     The current is  [tex]I = 6.5 *10^{-5} \ A[/tex]

     The time constant is  [tex]\tau = 4.8 \ s[/tex]

The resistance of resistor is mathematically evaluated as

      [tex]R = \frac{E}{I}[/tex]

substituting values

      [tex]R = \frac{ 110 }{ 6.5*10^{-5}}[/tex]

      [tex]R = 1.692*10^{6} \Omega[/tex]

The  capacitance of the capacitor is mathematically represented as

        [tex]C = \frac{\tau}{R}[/tex]

substituting values

       [tex]C = \frac{ 4.8}{ 1.692*10^{6}}[/tex]

       [tex]C = 2.837 *10^{-6} \ F[/tex]

Answer

fringe separation l distance between maxima decreases

Explanation:

Because the wavelength of blue light is smaller than that if red light

Answer:

A.21.3T

B.1.8x 10^6J/m^3

C.0.27x10^2J

D.6.6x10^-3H

Explanation:

Pls see attached file

(a) The drift speed of electrons in the wire is 2.22 x 10⁻⁴ m/s.

(b) The potential difference between two points in the wire is 0.013 V.

The given parameters;

The drift speed of electrons in the wire is calculated as follows;

[tex]v_d = \frac{I}{qn A} \\\\but , \ \frac{I}{A} = \frac{E}{\rho} \\\\v_d = \frac{E}{qn \rho}[/tex]

where;

[tex]v_d = \frac{0.052}{1.602 \times 10^{-19} \times 8.5\times 10^{28} \times 1.72 \times 10^{-8}} \\\\v_d = 2.22 \times 10^{-4} \ m/s[/tex]

The potential difference between two points in the wire, separated by 25 cm;

V = Ed

where;

V = 0.052 x 0.25

V = 0.013 V

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Answer:

Vx' = (Vx - u) / (1 - Vx *u / c^2)      velocity transformation formula

In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship

VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c

VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c

In both cases an observer on earth will observe the light traveling at speed c.

Answer:

53.3micro meters

Explanation:

See attached file

Answer:

Vf = 14.7 m/s

Explanation:

Vf² = Vi² + 2 * a * Δy

given:

a = 9.81 m/s²

Δy = 11m

Vi = 0 when upon release

Vf² = 0 + 2 (9.81) 11

Vf = 14.7 m/s

The velocity of the ball when it hits the ground will be 14.7 m/s.

The change of displacement with respect to time is defined as the velocity.  Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

given:

a(gravitational acceleration) = 9.81 m/s²

s (distance)= 11m

v is final velocity

u is the initial velocity

From Newton's second equation of motion;

[tex]\rm v^2 = u^2+2as \\\\ v^2=2 \times 9.81 \times 11 \\\\ v= 14.7 \ m/sec[/tex]

Hence, the velocity of the ball when it hits the ground will be 14.7 m/s.

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Answer:

Explanation:

Here potential energy lost = mgh

h = L ( 1 - cos 46 ) where L is length of rope

= 2.6 x ( 1 - cos 46 )

PE lost = 20 x 9.8 x 2.6 ( 1 - cos 46 )

= 155.6 J

gain of kinetic energy = loss of PE

1/2 m v² = 155.6

.5 x 20 x v² = 155.6

v² = 15.56

v = 3.94 m /s

Answer:

The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].

Explanation:

Let be an uniform rod of length L whose origin is located at one one end and axis is perpendicular to the rod, such that:

[tex]\lambda = \frac{dm}{dr}[/tex]

Where:

[tex]\lambda[/tex] - Linear density, measured in kilograms per meter.

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

[tex]r[/tex] - Distance of a point of the rod with respect to origin.

Mass differential can translated as:

[tex]dm = \lambda \cdot dr[/tex]

The equation of the moment of inertia is represented by the integral below:

[tex]I = \int\limits^{L}_{0} {r^{2}} \, dm[/tex]

[tex]I = \lambda \int\limits^{L}_{0} {r^{2}} \, dr[/tex]

[tex]I = \lambda \cdot \left(\frac{1}{3}\cdot L^{3} \right)[/tex]

[tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex] (as [tex]m = \lambda \cdot L[/tex])

The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].

Answer:

Explanation:

Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g

U is the initial velocity of the body (in m/s)

Ф is the angle of projection

g is the acceleration due to gravity.

Given U = 14m/s, g = 9.8m/s and range R = 15 m

we will substitute this value into the formula to get the projection angle Ф as shown;

15 = 15²sin2Ф/9.8

15*9.8 = 15²sin2Ф

147 = 225sin2Ф

sin2Ф = 147/225

sin2Ф = 0.6533

2Ф = sin⁻¹0.6533

2Ф = 40.79°

Ф = 40.79°/2

Ф = 20.39° ≈ 20°

Hence, the range is greatest at angle 20°

Answer:

The positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.

Explanation:

Given;

distance between the conducting planes, d = 80 cm

frequency of the electromagnetic wave, f = 750 MHz

speed of light, c = 3 x 10⁸ m/s²

Determine the wavelength

λ = C/f

where;

λ is the wavelength

C is the speed of light

f is the frequency

λ = C/f

λ = (3 x 10⁸) / (750 x 10⁶)

λ = 0.4 m = 40 cm

One complete cycle = one wavelength = 40 cm

half of the wavelength ( λ / 2) = 20 cm

one wavelength + half wavelength (3λ / 2) = 60 cm

The positions of the wave at zero amplitude (between 0 and 80 cm) = 20 cm, 40 cm, 60 cm

Thus, the positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.

Answer:

The dimension of mass can be represented as:  [tex][F^{1} T^{1} V^{-1} ][/tex]

Explanation:

We have  Force = Mass X Acceleration

                           = Mass X [tex]\frac{Change in Velocity}{Time Taken}[/tex]

                or, Mass  = Force x [tex]\frac {Time Taken } { Change in Velocity }[/tex]

                             So, dimensions of mass  = [tex]\frac{[F][T]}{[V]}[/tex]

                                                                       = [tex][F^{1} T^{1} V^{-1} ][/tex]

Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.

Answer:

f1 / f2 = n2 / n1  

Explanation:

To solve this problem, we should remember that the formula for index of refraction is defined as:

n = c / v

or

n v = c

Where,

n = index of refraction

c = speed of light

v = speed of light in the medium

Since speed of light is constant, then we can simply equate the materials 1 and 2:

n1 v1 = n2 v2

Where the speed of light in the medium (v) can be expressed as:

v = w * f

Where,

w = wavelength of light

f = frequency of light

Therefore substituting this back into the relating equation:

n1  w1 f1 = n1  w2 f1

Since it is given that the light is monochromatic, w1 = w2, this further simplifies the equation to:

n1 f1 = n2 f2

f1 / f2 = n2 / n1                  (ANSWER)

Complete question is;

Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun

Answer:

Minimum distance = 147,099,713.4 km

Maximum distance = 152,096,286.6 km

Explanation:

The formula for the eccentricity of an ellipse is given by;

e = c/a

where;

c is distance from the center of the ellipse to the focus of the ellipse.

a is distance from the center of ellipse to a vertex.

From the question, we are given;

a = 149,598,000

e = 0.0167

Thus;

0.0167 = c/149,598,000

c = 0.0167 × 149,598,000

c = 2,498,286.6

Now, formula for the minimum distance (perihelion) is;

Minimum distance = a - c

Minimum distance = 149598000 - 2498286.6

Thus;

Minimum distance = 147,099,713.4 km

Similarly, formula for the maximum distance (aphelion) is;

Max distance = a + c

Max distance = 149598000 + 2498286.6

Maximum distance = 152,096,286.6 km

The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

The given parameters;

The work done by the tension is calculated as follows;

W = Fd

W = 15 x 0.15

W = 2.25 J

The work done by gravity is calculated as;

W = (25 x 9.8) x 0.15

W = 36.75 J

Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

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Answer:

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

Explanation:

Consider a wire with:

Resistance = R

Length = L

Area = A = πr²

where, r = radius

ρ = resistivity

Then:

R = ρL/A

R = ρL/πr²   --------------- equation 1

Now, the new wire has:

Resistance = R'

Resistivity = ρ

Length = L' = 2 L

Radius = r' = 2r

Area = πr'² = π(2r)² = 4πr²

Therefore,

R' = ρL'/πr'²

R' = ρ(2 L)/4πr²

R' = (1/2)(ρL/πr²)

using equation 1:

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

Answer:

The approximate displacement of the object is  23  m.

Explanation:

Given that:

v = 4t + 5 (m/s)  for 3< t< 7; n= 4

The approximate displacement of the object can be calculated as follows:

The velocities at the intervals of t are :

3

4

5

6

the velocity at the intervals of t =  7 will be left out due the fact that we are calculating the left endpoint Reimann sum

n = 4 since there are 4 values for t, Then there is no need to divide the velocity values

v(3) = 4(3)+5

v(3) = 12+5

v(3) = 17

v(4)= 4(4)+5

v(4) = 16 + 5

v(4) = 21

v(5)= 4(5)+5

v(5) = 20 + 5

v(5) = 25

v(6) = 4(6)+5

v(6) = 24 + 5

v(6) = 29

Using Left end point;

[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]

= 23 m

Answer:

Explanation:

a )

current in the wire = potential diff / resistance

= 23 / (15 x 10⁻³ )

= 1.533 x 10³ A .

b )

For resistance of a wire , the formula is

R = ρ L / S where ρ is specific resistance , L is length and S is cross sectional area of wire

putting the given values

15 x 10⁻³ = 4ρ / π x .003²

ρ = 106  x 10⁻⁹ ohm. m

= 10.6 x 10⁻⁸ ohm m

The metal wire appears to be platinum .

(a) The current in the wire is 1.533 x 10³ A

(b) The resistivity of the wire material is 10.6 x 10⁻⁸ Ωm

(c) The material of the wire is Platinum

(a) According to the Ohm's Law:

V = IR

where V is the potential difference

I is the current

and R is the resistance

So,

I = V/R

I = 23 / (15 x 10⁻³ )

I = 1.533 x 10³ A

(b) The resistance of a wire is expressed as:

R = ρ L / A

where ρ is the resistivity,

L is length

and A is the cross-sectional area

15 x 10⁻³ = 4ρ / π x .003²

ρ = 106  x 10⁻⁹ Ωm

ρ = 10.6 x 10⁻⁸ Ωm

The metal from which the wire is made is platinum.

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Answer:

The quantities that always have equal magnitude for both masses during the collision are change in momentum of the colliding bodies and force exerted by each body

Explanation:

During collision of two bodies, the following quantities are affected;

Kinetic energy of the colliding bodies

change in momentum of the colliding bodies

Force exerted by each body

Two bodies that stick together after collision is inelastic collision.

For inelastic collision, the kinetic energy before collision is greater than kinetic energy after collision.

Change in momentum is zero, that is, momentum before collision is equal to momentum after collision.

According Newton's third law of motion, the force exerted by each body is equal but acts in opposite direction.

Therefore, the quantities that always have equal magnitude for both masses during the collision are change in momentum of the colliding bodies and force exerted by each body

The quantities that always have equal magnitude for both masses during the collision are change in momentum and force exerted by each body

Inelastic collision is a collision in which both bodies stick with each other after collision.

For inelastic collision, the momentum before collision is equal to momentum after collision.

Also, the force exerted by each body is equal but acts in opposite direction.

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Answer:

The drag coefficient of the car is 0.189

Explanation:

mass of the car = 2250 kg

Frontal area of the car  = 2.35 m^2

initial speed of the car = 72 km/hr = (72 x 1000)/3600 = 20 m/s

final speed of the car = 54 km/hr = (54 x 1000)/3600 = 15 m/s

time taken by the car to slow down = 105 sec

We'll assume that the value of the drag coefficient is constant throughout the deceleration.

The car decelerates from 20 m/s to 15 m/s in 105 seconds, the deceleration is calculated from

[tex]a = \frac{v-u}{t}[/tex]

where a is the deceleration

v is the final speed of the car

u is the initial speed of the car

t is the time taken to decelerate.

imputing values, we'll have

[tex]a = \frac{15-20}{105}[/tex] = -0.0476 m/s^2  (the -ve sign indicates a deceleration, which is a negative acceleration)

we can safely ignore the -ve sign in other calculations that follows

The force (drag force) with which the air around the decelerates the car is equal to..

[tex]F_{D} = ma[/tex]

where [tex]F_{D}[/tex] is the drag force

m is the mass of the car

a is the deceleration of the car

imputing values, we'll have

[tex]F_{D} = 2250*0.0476[/tex] = 107.1 N

equation for drag force is

[tex]F_{D} = \frac{1}{2}pAC_{D} v^{2}[/tex]

where p is the air density ≅ 1.225 kg/m³

A is the frontal area of the car

[tex]C_{D}[/tex] is drag coefficient of the car

v is the relative velocity of air and the car, and will be taken as the initial velocity of the car before starting to decelerate.

imputing these values, we'll have

[tex]107.1 = \frac{1}{2}*1.225*2.35*C_{D}*20^{2}[/tex] = 575.75[tex]C_{D}[/tex]

[tex]C_{D}[/tex] = 107.1/575.75 = 0.189

Answer:

Q = -3280J

Explanation:

From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:

ΔU = Q + W        ----------(i)

Where;

ΔU = total change in internal energy of a system.

Q = heat exchanged between the system and the surrounding

W = work done by or on the system.

If heat is lost into the surrounding, then Q = -ve, else Q = +ve

If work is done on the system, then W = -ve, else W = -ve

=> From the question, Rick is the system and does a work of

W = +1090J  [since Rick does the work, W = +ve]

=>Also, the internal energy decreases by 2190J, therefore,

ΔU = -2190J   [since there is a decrease in internal energy]

Substitute the values of W and ΔU into equation (i) as follows;

-2190 = Q + 1090

=> Q = -2190 - 1090

=> Q = -3280J

Therefore, the value of Q = -3280J

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:

6.8%

Explanation:

According to Stefan-Boltzmann law, radiation is directly proportional with temperature raised to the fourth power:

P ∝ T⁴

Writing a proportion:

P₁ / P₂ = T₁⁴ / T₂⁴

1.3P / P = (T₁ / T₂)⁴

T₁ / T₂ = ∜1.3

T₁ = 1.068 T₂

The temperature increased by 6.8%.