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3 years ago

6

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1 answer:

lidiya [134]3 years ago

8 0

3,728.2271 miles per hour please mark brainliest

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The amplitude of a paricular wave is 4.0 m. The crest to trough distance

kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0

3 years ago

Which of the following would have the most momentum?

Anarel [89]

I think it should be D as momentum is the product of mass and velocity...

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2 years ago

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference

strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

$n_{air}=1$

$\theta_{liquid} = 19.38\°$

$\theta_{air}35.09\°$

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

$n_1sin\theta_1 = n_2sin\theta_2$

$n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}$

$n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}$

Replacing the values we have:

$n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}$

$n_liquid = 1.7323$

Therefore the refractive index for the liquid is 1.7323

6 0

3 years ago

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

#A

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

#B

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

#C

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

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2 years ago

A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density ρ is not un

baherus [9]

A negative charge and a positive charge

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3 years ago