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egoroff_w [7]
3 years ago
6

which type of wave spreading do you think causes faster energy loss-two-dimensional or three-dimensional? explain.

Physics
2 answers:
madreJ [45]3 years ago
4 0

Answer:

Three dimensional wave

Explanation:

We know that waves generally spread in multiple dimensions and have different wavefronts, the more a wavefront moves away from the source it spreads the energy as it becomes larger hence it starts loosing energy.

As the number of wavefront in a three dimensional wave will certainly be more than that of two dimensional wave, it will spread more energy and will definitely starts to loose its energy way faster than two dimensional wave

sdas [7]3 years ago
3 0
Three dimensional would loose faster 

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At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d =
defon

Answer:

6.03 mV

Explanation:

length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,

Magnetic field due to solenoid

B = μ0 n i = μ0 N i / L

dB/dt = μ0 N / L x di / dt

dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s

Induced emf, e = rate of change of magnetic flux

e = dΦ / dt = A x dB / dt

e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV

7 0
2 years ago
A potter’s wheel moves from rest to an angular speed of 0.50 rev/s in 28.9 s. Assuming constant angular acceleration, what is it
Fudgin [204]

Answer:

0.108 rad/s².

Explanation:

Given that

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 0.5 rev/s

We know that

1 rev/s = 6.28 rad/s

ωf= 3.14 rad/s

t= 28.9 s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

3.14 = 0 + α x 28.9

\alpha=\dfrac{3.14}{28.9}\ rad/s^2\\\alpha=0.108\ rad/s^2

Therefore the acceleration will be 0.108 rad/s².

Therefore the answer will be 0.108 rad/s².

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