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3 years ago

Sb-28 a collision could occur when the distance decreases and bearing between two vessels does what?

Physics

1 answer:

USPshnik [31]3 years ago

4 0

A collision happens when the distance between two boats decreases and the bearings aren't changing.

To avoid collisions, the operators should master the Rules of the Road. The operators should be able to determine if his boat is a Stand-On Vessel (maintains course and speed) or the give-way vessel (the "burdened" vessel which gives way to the stand-on vessels),and maneuver accordingly.

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Chemists use a wide array of techniques for determining the exact composition and structure of a compound. One of the most robus

attashe74 [19]

Answer:

Δμ = hΔf/B

Explanation:

If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.

Since the magnetic energy change equals the photon energy,

ΔE = ΔU

hΔf = ΔμB

Δμ = hΔf/B

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3 years ago

Write a script for a 5-min mock debate between Edison and Tesla, in which each tries to convince the audience that he was the mo

myrzilka [38]

Answer:

edieson and atela script once upon a time there was the name is a copy paste and the you can come by the office and I have a car so we can do it tomorrow if I will be working from the gathered that was a week if we were in w

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3 years ago

Can someone pleaseEeeeeeee ASAP answer this❗️❗️❗️❗️ I really need help it is my second time posting this❗️❗️❗️❗️

belka [17]

Answer:

Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?

Explanation:

So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

$\sqrt{t } = \sqrt{77.55}$

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull

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3 years ago

How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec

Rus_ich [418]

Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

$\text{ Initial vertical velocity = 0 m/s}$

let's find how long the ball remained in the air.

$\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}$

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

$\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}$

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1 year ago

What causes a regular reflection

bezimeni [28]

Regular reflection is a reflection such that the the angle of reflection of the light is equal to the angle of incidence and it is on the opposite of the normal to the point of incidence. In a regular reflection the incident ray, the reflected ray and the normal to the reflection all lie in the same plane. Regular reflection happens if the reflection surface is very smooth.

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3 years ago