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Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—than humans. What are the two lowest frequencies at which dogs have an increase in sensitivity? The speed of sound in the warm air of the ear is 350 m/s.A. 1700 Hz, 3400 Hz
B. 1700 Hz, 5100 Hz
C. 3400 Hz, 6800 Hz
D. 3400 Hz, 10,200 Hz

Answers

Answer 1

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

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A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.20 ?

Answers

Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

Time, t = 1 s

resistance of the coil, R = 1.2 ohm

The induced emf is given by

$e=(d\phi)/(dt)=A(B' - B)/(t)$

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

i = e/R

i = 2.184/1.2

i = 1.82 mA

A wheel 2.45 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.30 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.a. the angular speed of the wheel and, for point P
b. the tangential speed.
c. the total acceleration.
d. the angular position.

Answers

(a) The angular speed of the wheel at point P is 8.6 rad/s.

(b) The tangential speed of the wheel is 10.54 m/s.

(d) The angular position of the wheel is 87 ⁰.

The given parameters;

diameter of the wheel, d = 2.45 m
radius of the wheel, r = 1.225 m
angular acceleration of the wheel, α = 4.3 rad/s²
angular displacement of the wheel, θ = 57.3⁰
time of motion, t = 2.0 s

The angular speed of the wheel at point P is calculated as follows;

$\omega_f = \omega _i + \alpha t\n\n\omega _f = 0 + 4.3 * 2\n\n\omega _f = 8.6 \ rad/s$

The tangential speed of the wheel is calculated as follows;

$v = \omega _f r\n\nv = 8.6 * 1.225 \n\nv = 10.54 \ m/s$

The centripetal acceleration of the wheel is calculated as follows;

$a_c = (v^2)/(r) \n\na_c = ((10.54)^2)/(1.225) \n\na_c = 90.69 \ m/s^2$

The total acceleration of the wheel is calculated as follows;

$a_t = √(a_c^2 + a_r^2) \n\na_t = √(90.69^2 + 4.3^2) \n\na_t = 90.8 \ m/s^2$

The angular position is calculated as follows;

$\theta = tan^(-1) ((a_c)/(a_r) )\n\n\theta = tan^(-1) ((90.69)/(4.3) )\n\n\theta = 87 \ ^0$

Learn more here: brainly.com/question/14508449

Answer:

Explanation:

Radius of wheel R = 1.225 m

For angular motion of wheel

ω = ω ₀ + α t

= 0 + 4.3 x 2

= 8.6 rad / s

This is angular speed of wheel and point P .

b )

Tangential speed = ωR

8.6 x 1.225

= 10.535 m / s

c )

radial acceleration

a_r = v² / r

= 10.535² / 1.225

= 90.6 m / s²

tangential acceleration = radius x angular acceleration

a_t = 4.3 x 1.225

= 5.2675

Total acceleration = √ 90.6² + 5.2675²

= √ 8208.36 + 27.7465

= 90.75 m/s²

d ) angle of rotation

= 1/2 α t²

= .5 x 4.3 x 4

= 8.6 radian

= (8.6/3.14) x 180

= 499 degree

= 499 + 57.3

= 556.3

556.3 - 360

= 196.3 degree

Point p will rotate by 196.3 degree

Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. What is the ratio of the radius of A to that of B

Answers

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

FOR WIRE A:

R₁ = ρ₁L₁/πr₁² -------- equation 1

FOR WIRE B:

R₂ = ρ₂L₂/πr₂² -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

r₁/r₂ = 1/2 = 0.5

Final answer:

The ratio of the radius of wire A to the radius of wire B is 1/2.

Explanation:

The resistance of a wire is given by the formula R = ρl/A, where R is resistance, ρ is resistivity, l is length, and A is the cross-sectional area of the wire. When the wire has a circular cross-section, the area can be calculated by the formula A = πr². The resistance of the wire then becomes: R = ρl/(πr²). If the resistance of wire A is four times that of wire B, we can set up the equation 4RB = RA. Substituting the expression for resistance, we get 4(ρl/(πrB²)) = ρl/(πrA²). Simplifying, we find that the ratio of the radius of wire A to the radius of wire B is one-half, or rA/rB = 1/2.