On the graph below, the line crosses the x axis on a velocity versus time

Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resulta

andreev551 [17]

Answer:

Therefore the amplitude of the resultant wave is $=0.95 y_m$

Explanation:

The equation of wave:

y=A sin (kx-ωt)

For wave 1:

y₁=A sin (kx-ωt) = $y_{m}$ sin (kx-ωt)

For wave 2:

y₂=A sin (kx-ωt+Φ) = $y_{m}$ sin (kx-ωt+Φ)

Where A= amplitude= $y_m$

The angular frequency $\omega=\frac{2\pi}{T}$

$k=\frac{2\pi}{\lambda}$ , $\lambda$ = wave length.

t= time

T= Time period

$\phi$ = phase difference = $\frac{\pi}{5}$

The resultant wave will be

y = y₁ + y₂

= $y_m$ sin (kx-ωt) + $y_m$ sin (kx-ωt+Φ)

$=y_m$ {sin (kx-ωt) + sin (kx-ωt+Φ)}

$=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)$

$=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)$

Therefore the amplitude of the resultant wave is

$=y_m\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\pi }{10})$

$=0.95 y_m$

6 0

4 years ago

Which of the following statements concerning the nuclear force is false? O The nuclear force is attractive and not repulsive. O

Ulleksa [173]

Answer:

The nuclear force is attractive and not repulsive.
The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces.

Explanation:

Nuclear force is the strongest existing force in the nature.

It has the shortest range.

Its main function is to hold the subatomic particles together in nature.

The nuclear force is created by the exchange of pi mesons between the nucleons of an atom, but for this exchange to happen the particles must be close to one another of the order of few femtometer.

At about 1 femtometer the nuclear force is very strongly attractive in nature but at distance greater than 2.5 femtometer it fades away.

The force becomes repulsive in nature at distance less than 0.7 femtometer.
This force holds the likely charged protons together in the nucleus.

3 0

3 years ago

Scientists observe an approaching asteroid that is on a collision course with

nasty-shy [4]

Answer:

The approximate velocity the rocket must have to stop the asteroid completely after the collision is;

C. -324 m/s

Explanation:

The parameters of the asteroid and the rocket are;

The mass of the asteroid, m₁ = 11,000 kg

The initial velocity with which the asteroid is approaching Earth, v₁ = 50 m/s

The mass of the rocket, m₂ = 1700 kg

The initial velocity of the rocket = v₂

The final velocity of the combined asteroid and rocket after the collision, v₃ = 0 m/s

By the law of conservation of linear momentum, we have;

The total initial momentum = The total final momentum

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Substituting the known values, we get;

11,000 kg × 50 m/s + 1,700 kg × v₂ = (11,000 kg + 1,700 kg) × 0 m/s

11,000 kg × 50 m/s + 1,700 kg × v₂ = 0

∴ 1,700 kg × v₂ = -11,000 kg × 50 m/s

v₂ = (-11,000 kg × 50 m/s)/(1,700 kg) = -323.529412 m/s ≈ -324 m/s

The approximate initial velocity the jet must have to completely stop the asteroid after the collision is -324 m/s.

3 0

3 years ago

A very long thin wire carries a uniformly distributed charge, which creates an electric field. The electric field is (2300 N/C ,

Leto [7]

Answer:

λ= 5.24 × 10 ⁻² nC/cm

Explanation:

Given:

distance r = 4.10 cm = 0.041 m

Electric field intensity E = 2300 N/C

K = 9 x 10 ⁹ Nm²/C

To find λ = linear charge density = ?

Sol:

we know that E= 2Kλ / r

⇒ λ = -E r/2K (-ve sign show the direction toward the wire)

λ = (- 2300 N/C × 0.041 m) / 2 × 9 x 10 ⁹ Nm²/C

λ = 5.24 × 10 ⁻⁹ C/m

λ = 5.24 nC/m = 5.24 nC/100 cm

λ= 5.24 × 10 ⁻² nC/cm

3 0

3 years ago

Calculate the average times it took the car to travel 0. 25 and 0. 50 meters. Record the averages, to two decimal places, in Tab

Illusion [34]

The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

$S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22$

For the 0.50 m, the average speed will:

$S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg} = 3.13\rm \ s$

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

Learn more about Average speed:

brainly.com/question/26386984

6 0

3 years ago