Which of the following is NOT a fundamental S.I unit? A. Meter B. Ampere C. Kelvin D. Radian

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

3 years ago

A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi

Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0

3 years ago

A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the

lilavasa [31]

Answer:

$R=803k\Omega$

Explanation:

We have the following information,

$V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s$

We apply the equation for capacitor charging the voltage across it,

$V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}$

Replacing values,

$R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}$

$R=803k\Omega$

3 0

3 years ago

A greenhouse is an example of a(n) __________________________________. Question 7 options: active solar energy system. biomass g

Ray Of Light [21]

An Active solar energy system because it uses the sun to heat up the green house an I’m return the plants grow an produce food

8 0

2 years ago