friction always opposes the _____​

This question is incomplete, the complete question is;

A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answer:

the second moth is moving at 5.062 m/s

Explanation:

Given the data in the question;

Using doppler's effect

[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )

f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )

frequency reflected from the moth,

Now, moth is the source and the bat is the receiver

f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )

hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )

we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.

given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.

we substitute

55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[  ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )

1.02996 =  ( 343 + u₂ ) / ( 343 - u )

( 343 + u₂ ) = 1.02996( 343 - u )

343 + u = 353.27628 - 1.02996u

u + 1.02996u = 353.27628 - 343

2.02996u = 10.27628

u = 10.27628 / 2.02996

u = 5.062 m/s

Therefore, the second moth is moving at 5.062 m/s

Answer:

the required angle is 0.0834879⁰

Explanation:

Given  the data in the question;

slit separation; d = 1.75 mm = 1.75 × 10⁻³ m

wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m

wavelength λ₂ 510 nm = 510 × 10⁻⁹ m

Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;

tanθ = [tex]y_m[/tex] / D = mλ/d

since they both coincides;

tanθ₁ = tanθ₂

m₁λ₁/d = m₂λ₂/d

multiply both sides by d

so,

m₁/m₂ = λ₂/λ₁

we substitute

m₁/m₂ = 510 nm / 425 nm

m₁/m₂ = 510 nm / 425 nm

divide through by 85

m₁/m₂ = 6 / 5

hence m₁ and m₂ are 6 and 5

so, from the previous formula

tanθ₂ = m₂λ₂/d

we substitute

tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m

tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m

tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m

tanθ₂ = 0.00145714

θ₂ = tan⁻¹( 0.00145714 )

θ₂ = 0.0834879⁰

Therefore, the required angle is 0.0834879⁰

Answer:

şen çal kapimi turkish drama

Answer:

|x| = √53

Explanation:

We are told that the vector starts at the point (0.0) and ends at (2,-7) .

Thus, magnitude of displacement is;

|x| = √(((-7) - 0)² + (2 - 0)²)

|x| = √(49 + 4)

|x| = √53

Answer:

Standard units are commonly used units of measurement, which help us measure length, height, weight, temperature, mass and more. These units are standardised, which means that everyone gets the same understanding of the size, weight and other properties of objects and things.

Explanation:

The minimum width of the screen is 34 cm.

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m

So, sinθ = mλ/d

Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.

For small angles sinθ ≈ tanθ

So, mλ/d = L/D

making L subject of the formula, we have

L = mλD/d

L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷  1/5640 × 10⁻² m per line

L = 1 × 455 × 10⁻⁹ m × 0.661 m  × 5640 × 10² line per m

L = 1696258.2 × 10⁻⁷ m

L = 0.16963 m

L ≅ 0.17 m

So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.

So, w = 2 × 0.17 m

w = 0.34 m

w = 34 cm

So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.

Learn more about diffraction grating here:

https://brainly.com/question/15712101

Being in parallel each device will have an equal voltage drop of 120 V

A. Yes the combination will blow the fuse. See part B for the total current.

B. Toaster = 1800W / 120V = 15A

   Frying Pan = 1400W / 120V = 11.67A

  Lamp = 55W / 120V = 0.458A

Total amps = 15 + 11.67 + 0.458 = 27.128 Amps

27.128A is greater than 15A so the fuse will blow.

Answer:

-4/7

Explanation:

Given the following

A=4i-10j and B= 7i+5j

A+ bB = 4i-10j + (7i+5j)b

A+ bB =  4i-10j + 7ib+5jb

A+ bB =

The vector along the x-axis is expressed as i + 0j

If the vector A+ bB is pointing in the direction of the x-axis then;

[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]

Hence the value of b is -4/7

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

According to the statement, we have following system of vectorial equations:

[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)

[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)

[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)

By applying (1) and (2) in (3):

[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]

[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]

And we get two scalar equations after analyzing each component:

[tex]4+7\cdot \beta = c[/tex] (4)

[tex]-10+5\cdot \beta = 0[/tex] (5)

We solve for [tex]\beta[/tex] in (5):

[tex]\beta = 2[/tex]

And for [tex]c[/tex] in (4):

[tex]c = 4+7\cdot (2)[/tex]

[tex]c = 18[/tex]

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720

Explanation:

Given that,

In order to find the distance travelled, firstly we need calculate the acceleration.

→ v = u + at

→ 20 = 0 + 60a

→ 20 = 60a

→ 20 ÷ 60 = a

→ ⅓ m/s² = a

Now, by using the 2nd equation of motion :

→ s = ut + ½at²

→ s = 0(60) + ½ × ⅓ × (60)²

→ s = ⅙ × 3600

→ s = 1 × 600

→ s = 600 m

Hence, the distance travelled is 600 m.

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

Answer:

work done is -2.8  × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

[tex]W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta[/tex]

[tex]W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta[/tex]

[tex]W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta[/tex]

W = [tex]-mgl[[/tex] -cosθ [tex]]^{0.047}_{0.045 }[/tex]

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8  × 10⁻⁶ J

Therefore, work done is -2.8  × 10⁻⁶ J

Answer:

[tex]R=0.023m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.16*10^{-26}[/tex]

Potential difference [tex]V=523V[/tex]

Magnitude [tex]m=0.370 T[/tex]

Generally the equation for Velocity is mathematically given by

[tex]\frac{1}{2}mv^2=ev[/tex]

[tex]v=\frac{2ev}{m}[/tex]

[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]

[tex]v=12.22*10^4m/s[/tex]

Generally the equation for Force is mathematically given by

[tex]F=qvBsin \theta[/tex]

Where

[tex]qVB=m\frac{v^2}{R}[/tex]

[tex]F=m\frac{v^2}{R}sin\theta[/tex]

Therefore

[tex]R=\frac{mv}{qB sin \theta}[/tex]

[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]

[tex]R=0.023m[/tex]

Answer:

56.1 kg

Explanation:

Given

[tex]T = 2200Nm[/tex] -- torque

[tex]d_1 = 1.0m[/tex]

[tex]d_2 = 3.0m[/tex]

Required

The mass of the diver

From the question, we understand that the diver is at the extreme of the board.

So, we make use of the following torque equation

[tex]T = F * (d_1 + d_2)[/tex]

Where:

[tex]F \to Force[/tex]

So, we have:

[tex]2200 = F * (1.0 + 3.0)[/tex]

[tex]2200 = F * 4.0[/tex]

Divide both sides by 4.0

[tex]550 = F[/tex]

[tex]F = 550 N[/tex] --- This is the force exerted by the diver (in other words, the weight).

To calculate the mass, we use:

[tex]F = mg[/tex]

Make m the subject

[tex]m = \frac{F}{g}[/tex]

This gives:

[tex]m = \frac{550}{9.8}[/tex]

[tex]m = 56.1kg[/tex]

Answer:

the answer is 49000 joules at the maximum height

Explanation:

we know the mass (50kg)

we know the acceleration due to gravity(9.8m/s²)

we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m

the formula is potential energy=mgh

m for mass

g for acceleration due to gravity

h for height

multiply them

50x9.8x100

we get 49000

the unit of potential energy is joules so the answer is

49000 joules

Answer:

49000 joules

Explanation:

hope it helpss

Answer:

The answer would be D), if the decay is beta negative.

Explanation:

Thorium-232 goes through alpha decay:

Thorium-232 --> Helium-4 + Radium-228

Radium-228 then can undergo beta positive or beta negative decay:

Beta positive = Radium-228 --> Electron + Francium-228

Beta negative = Radium-228 --> Positron + Actinium-228

Therefore, the isotope that is created is Actinium-228

Answer:

Types of tidal turbines

Axial turbines.

Crossflow turbines.

Flow augmented turbines.

Oscillating devices.

Venturi effect.

Tidal kite turbines.

Turbine power.

Resource assessment.

Answer:

Axial turbines

Crossflow turbines

flow augmented turbines

Answer:

the force happens on the wall and couch

Explanation:

she is using her arm strength to lift and hold

Answer:

The angular acceleration is 26.6 rad/s^2.

Explanation:

Force, F = 314 N

radius, r = 0.281 m

mass, m = 84.2 kg

The grindstone is a disc.

The torque is given by

torque = force x radius

Torque = 314 x 0.281 = 88.234 Nm

The torque is given by

Torque = Moment of inertia x angular acceleration

[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]

Answer:

[tex]v=(6ti+6k)\ m/s[/tex]

Explanation:

Given that,

The position of a particle is given by :

[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]

Let us assume we need to find its velocity.

We know that,

[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]

So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].

Answer:

v = 3.12 m/s

Explanation:

First, we will find the length of the string by using the formula of the time period:

[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]

where,

l = length of string = ?

T = time period = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]

Now, we will find tension in the string in the vertical position through the weight of the ball:

T = W = mg = (3 kg)(9.81 m/s²)

T = 29.43 N

Now, the speed of the transverse wave is given as follows:

[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]

v = 3.12 m/s

Answer:

the radius of the bigger loop is 5 cm.

Explanation:

Given;

current in the smaller loop, I₁ = 12 A

current in the larger loop, I₂ = 20 A

radius of the smaller loop, r₁ = 3 cm

let the radius of the larger loop, = r₂

Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.

[tex]B= \frac{\mu_0 I}{2r}[/tex]

The magnetic field at the center of the smaller loop;

[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]

The magnetic field at the center of the bigger loop;

[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]

If the magnetic field at the center is zero, then B₁ = B₂

[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]

Therefore, the radius of the bigger loop is 5 cm.

Answer:

F' = 128 N

Explanation:

The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:

[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]

where,

F = Force of attraction = 1 N

G = universal gravitational constant

q₁ = magnitude of the first charge

q₂ = magnitude of the second charge

r = distance between charges

Therefore,

[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)

Now, we apply the changes given in the question:

[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]

using eq (1):

F' = 128(1 N)

F' = 128 N

Answer:

[tex]N_f=248N[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=100kg[/tex]

Ladder Length [tex]l=4.0m[/tex]

Mass of Ladder [tex]m_l=25kg[/tex]

Angle [tex]\theta=56 \textdegree[/tex]

Generally the equation for Co planar forces is mathematically given by

[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]

Therefore

[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]

[tex]N_f=248N[/tex]

Answer:

질문?

Explanation:

평균 공기 밀도가 1.25kg/m³이고 수은 밀도가 13600kg/m³이고 g=10N/kg인 경우 산 기슭의 기압은 수은의 75.0cm이고 정상의 수은은 60cm입니다. 산의 높이를 계산?

68.6 bags are required to fill the specified volume of 10"x20"x20".

It is 40 lbs/1.9 gallons, assuming US gallons, for the topsoil.

US gallons equal 3785 ml. 1 lb = 1/2.2 = 0.454 kg or 454 grammes since 1 kilogramme equals 2.2 lbs. 3785 x 1.9 = 7192 ml is equal to 1.9 gallons. 40 lbs = 18160 g. Therefore, the topsoil density is 2.52 g/cc (18160/7192).

The volume of the peat bag is 14x20x30 inches, which is equal to 35.6x50.8x76.2 cm (1 inch = 2/54 cm)=137806 ccs.

In other words, for the 40 lb again, 18160 grams/137806 equals 0.13 g/cc. The peat is therefore significantly lighter than topsoil.

The volume of the latter volume is 120"x240"x20" or 576,000 cubic inches, and the volume of a bag is 14"x20"x30" or 8400 cubic inches, hence 576,000/8400 = 68.6 bags are required to fill the specified volume of 10"x20"x20".

Learn more about density herebrainly.com/question/4970627

SPJ1

Answer:

Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.

Explanation:

Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

Answer:

Two Main Branches of Physics

it is Classical Physics and Modern Physics.

Explanation:

Further sub Physics branches are Mechanics, Electromagnetism, Thermodynamics, Optics, etc. The rapid progress in science during recent years has become possible due to discoveries and inventions in the field of physics.

hope it helped