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zvonat [6]
4 years ago
10

If you start with two sinusoidal waves of the same amplitude traveling in phase on a string and then somehow phase-shift one of

them by 5.4 wavelengths, what type of interference will occur on the string?
Physics
1 answer:
andrew11 [14]4 years ago
7 0

Explanation:

Two waves with the same wavelength are in phase if there phase difference is zero or an integral multiple of wavelength. Thus the integer part of any difference expressed in wavelength can be discarded.

The phase difference of 5.4 wavelength is equivalent to is one of 0.4 wavelength thus the interference is an intermediate interference close to fully destructive.

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A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
lesya [120]

Answer:

Impulse in the first collision is greater

Explanation:

u = Initial velocity of cart

v = Final velocity of cart

m = Mass of of cart

First collision

Impulse

J=m(v-u)\\\Rightarrow J=m(0.3-(-0.3))\\\Rightarrow J=m(0.6)

Second collision

Impulse

J=m(v-u)\\\Rightarrow J=m(0-(-0.3))\\\Rightarrow J=m(0.3)

Hence, the impulse in the first collision is greater than the second impact

7 0
4 years ago
10 points and brainlyest if possible and right.
Delicious77 [7]
THE answer the table is holding the book up would be a normal force 
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4 years ago
Two rockets are fired at each other with initial velocities of 150m/s150m/s and are 6000m6000m apart. The first rocket is accele
Nataliya [291]

Answer:

3469.788 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

First rocket

s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m

Second rocket

s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m

When this will collide the total distance they would have covered would be 6000 m.

6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0

t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72

Hence at 13.72 seconds they will collide assuming they are launched at the same time.

s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m

The second rocket would have gone 3469.788 m when they collide

7 0
4 years ago
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Fed [463]

Answer:

1) you could put more force behind it. (increase) 2) have another object interact with that object. (increase or decrease) 3) the object could hit a wall and stop or slow down (decrease)

Sorry if wrong

5 0
3 years ago
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IrinaVladis [17]

Answer:

E

Explanation:

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3 years ago
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