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3 years ago

5

Is polarization possible for longitudinal waves ?

1 answer:

RideAnS [48]3 years ago

7 0

<span>"Longitudinal" is a polarization.A massive spin-1 boson has three possible polarizations: two transverse (like a photon), and one longitudinal.For purely longitudinal waves, such as sound waves in air, which cannot be transverse, we don't speak of polarization.</span>

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Water is flowing through a channel that is 12m wide with a

Alexus [3.1K]

Answer:

Velocity from second channel will be 1.6875 m/sec

Explanation:

We have given width of the channel , that is diameter of the channel 1 $d_1$ = 12 m

So radius $r_1=\frac{d_1}{2}=\frac{12}{2}=6m$

Speed through the channel 1 $v_1=0.75m/sec$

Width , that is diameter of the channel 2 $d_2=4m$

So $r_2=2m$

From continuity equation

$A_1v_1=A_2v_2$

$\pi \times 12^2\times 0.75=4\times \pi\times 4^2\times v_2$

$v_2=1.6875m/sec$

So velocity from smaller channel will be 1.6875 m /sec

6 0

3 years ago

Based on the data collected, explain why a launch angle of 30 degrees (HIGH HORIZONTAL VELOCITY & SMALL TIME OF FLIGHT) will

Gennadij [26K]

Answer:

Explanation:

The horizontal distance traveled by the projectile is given by the formula

$R=\frac{u^{2}Sin2\theta }{g}$

The formula for the time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Case I: when the launch angle is 30°

So, $R_{1}=\frac{u^{2}Sin60 }{g}$

$R_{1}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 30 = 0.866 u

$T_{1} = \frac{2u Sin30 }{g}=\frac{u}{g}$

Case II: when the launch angle is 60°

$R_{2}=\frac{u^{2}Sin120}{g}$

$R_{2}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 60 = 0.5 u

$T_{1} = \frac{2u Sin60 }{g}=\frac{1.73 u}{g}$

By observing the case I and case II, we conclude that

R1 = R2

Horizontal velocity 1 > Horizontal velocity 2

T1 < T2

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3 years ago

A student collects data on spring and neap tides, in feet, on four different days of the same year in Anchorage, Alaska. The gra

BabaBlast [244]

Answer:

September 9 and 24 represent spring tides due to the added gravitational pull of the Sun.

Explanation:

Just trust me

6 0

4 years ago

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

nignag [31]

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0

3 years ago

Please help, I really don't know

zubka84 [21]

circular motion.

cent acc = r omega^2 ... omega is ang vel ... omega=2pi/T ,,,

9.8=rx(2pi/T)^2

if r is known, solve for T

7 0

4 years ago