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3 years ago

Describe the responses of the human ear to sound waves coming from the

1 answer:

8 0

If the sound comes from the right side, the waves reach the right ear before the left ear. if the sound comes from the left side, the waves reach the left ear before the right ear. The difference between the phases of waves reaching both ears is detected by the ears and then interpreted by the brain

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec

borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

$\lambda = \frac{c}{f}$

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

$f = 2Hz$

$c = 3*10^8km/s$

Replacing we have,

$\lambda = \frac{c}{f}$

$\lambda = \frac{3*10^8km/s}{2Hz}$

$\lambda = 1.5*10^8m$

<em>Therefore the wavelength of this wave is $1.5*10^{8}m$ </em>

8 0

3 years ago

A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r

sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana $m_{b}$ = 8.97 kg

at distance r = 4/5 { radius of platform }

mass of monkey $m_{m}$ = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + $m_{b}$ ( $\frac{4}{5}$ R)² + $m_{m}$ R² ]w

m/2 × ω₀ = [ m/2 + $m_{b}$ ( $\frac{4}{5}$ )² + $m_{m}$ ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0

3 years ago

A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a

Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial Velocity of Each object

$V_f$ = Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0

3 years ago

A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?

Olenka [21]

Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
$F = k*x$

Solving:
$F = k*x$
$F = 50*0.15$
$\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark$

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
$E = \frac{k*x^2}{2}$

Solving:(Energy associated with this stretching)
$E = \frac{k*x^2}{2}$
$E = \frac{50*0.15^2}{2}$
$E = \frac{50*0.0225}{2}$
$E = \frac{1.125}{2}$
$\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago

Two disks of polaroid are aligned so that they polarize light in the same plane. Calculate the angle through which one sheet nee

Olegator [25]

Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle $\theta$ with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of $\cos^2\theta$ (this is the Law of Malus).

Explanation: Let us say we have a beam of unpolarized light of intensity $I_0$ that passes through two parallel Polaroid discs with the angle of $\theta$ between their planes of polarization. We are asked to find $\theta$ such that the intensity of the outgoing beam is $I_2$ . To solve this we follow the steps below:

Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

$I_1=\frac{I_0}{2}.$

This beam also becomes polarized in the plane of the first polaroid.

Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle $\theta$ with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of $\cos^2\theta.$ This yields $I_2=I_1\cos^2\theta$ . Substituting from the previous step we get

$I_2=\frac{I_0}{2}\cos^2\theta$

yielding

$\frac{2I_2}{I_0}=\cos^2\theta$

and finally,

$\theta=\arccos\sqrt{\frac{2I_2}{I_0}}$

3 0

3 years ago