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2 years ago

14

Suppose two waves collide, and the temporary combined wave that results is smaller than the original waves. what term best descr

ibes this interaction? (1 point diffraction destructive interference standing wave formation constructive interference

1 answer:

BlackZzzverrR [31]2 years ago

3 0

The collision of the two waves would bring force and that would make the new wave smaller

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PLEASE HELP ASAP WILL GIVE BRAINLIEST

stealth61 [152]

Explanation

(m) is measured in kilograms (kg)

<h2>(F) is measured in newtons (N)</h2>

<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>

4 0

2 years ago

Read 2 more answers

The orbit of the planets in our solar system is generally due to which characteristic of the Sun?

larisa [96]

The answer to your question is C. <span> the Sun's strong gravitational field . This is correct because i took the test :D</span>

6 0

3 years ago

Read 2 more answers

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat

Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, $v_{max}$ = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2$

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s$

Therefore, the maximum speed is 21.39 m/s.

3 0

2 years ago

A 1.5-kilogram cart initially moves at 2.0 m/s. It is brought to rest by a constant net force in 0.3 seconds. What is the magnit

ser-zykov [4K]

We know, F = m.a
Here, m = 1.5 Kg
a = v/t = 2/0.3 = 6.67 m/s

Substitute their values,
F = 1.5 * 6.67
F = 10 N

In short, Your Answer would be 10 Newtons

Hope this helps!

5 0

3 years ago