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1 year ago

A visible violet light emits light with a wavelength of 4.00 × 10-7 m.

Physics

2 answers:

RideAnS [48]1 year ago

4 0

Answer:

The correct option is B. 7.5 * 10¹⁴ Hz

Explanation:

Frequency = (speed) / (wavelength)

= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)

= (3/4 x 10¹⁵) ( m / m - s )

= (0.75 x 10¹⁵) /sec

= 7.5 x 10¹⁴ Hz

= 750,000 GHz

ludmilkaskok [199]1 year ago

3 0

Answer:

Mark Brainliest please

answer is

Explanation:

For any wave,

Frequency = (speed) / (wavelength)

= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)

= (3/4 x 10¹⁵) ( m / m - s )

= (0.75 x 10¹⁵) /sec

= 7.5 x 10¹⁴ Hz

= 750,000 GH

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A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.

nirvana33 [79]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

$v = r\omega \rightarrow \omega = \frac{v}{r}$

Here,

v = Lineal velocity

$\omega$ = Angular velocity

r = Radius

Our values are

$v = 6/ms$

$r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m$

Replacing to find the angular velocity we have,

$\omega = \frac{6m/s}{0.06m}$

$\omega = 100rad/s$

Convert the units to RPM we have that

$\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})$

$\omega = 955.41rpm$

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3 years ago

Which of the following is always true about the particles that make up matter? A. They can only be found in solid substances. B.

Neko [114]

B.They are too small to be seen with only our eyes

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2 years ago

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A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle

Artist 52 [7]

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

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2 years ago

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

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$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

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Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

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2 years ago

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