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Calculate the speed (in m/sec) of a wave with a wavelength of 2.1 meters and a period of 9.4 second.

Answers

Answer 1

Answer:

Wave speed = (wavelength) x (frequency)

We know the wavelength, but we don't know the frequency. How can we find the frequency ? "Here frequency frequency."

We know the period, and frequency is just (1 / period). So . . .

Wave speed = (wavelength) / (period)

Wave speed = (2.1 meters) / (9.4 seconds)

Wave speed = (2.1 / 9.4) m/s

Wave speed = 0.223 m/s

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A foot is 12 inches and a mile is 5280I ft exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3tall. what is Sammy's height in inches?

Answers

Answer:

65.3 Inches tall

Explanation:

If Sammy is 5 feet and 5.3 inches tall, we simply need to convert the feet to inches, and sum the remaining inches from his height to determine his overall height in inches.

So, 5 feet = (12 inches/1foot) * (5 feet) = 60 inches

And 60 inches + 5.3 inches = 65.3 inches.

Hence, Sammy is 65.3 inches tall.

Cheers.

A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse through a high-powered telescope he owns. You are curious what the eclipse might look like from different perspectives in space. If the moon has a diameter of 2,159.14 miles, what is the maximum distance that it could be observed by the naked eye with enough detail that you could distinguish it from other celestial bodies (assuming that you have 20/20 vision)

Answers

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

θ = 1.22 550 10⁻⁹ / 0.002

θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

tan θ = y / L

y = L tan θ

y = 2,389 10⁵ tan 3,355 10⁻⁴

y = 8.02 10¹ mi

y = 80.2 mille

This is the smallest size of an object seen directly by the eye

Final answer:

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

Explanation:

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

Learn more about Observing Moon here:

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An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball

Answers

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, $\omega=0.7\ rev/s=4.39\ rad/s$

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

$v=r\omega\n\nv=0.9* 4.39\n\nv=3.951\ m/s$

So, the tangential speed of the ball is 3.951 m/s.

How does activity on the Sun affect natural phenomena on Earth?

Answers

Answer and Explanation:

The Sun is the main source of energy on the earth if there will be no availability of Sun energy then life is impossible om the earth besides this the Sun warms our planet. The heating of ocean and atmosphere is mainly sue to Sun energy .Sun has also a great impact on the weather we can say that Sun is weather deciding on the earth our climate is totally dependent on the how much energy we got in form of radiation from earth.

7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, whatforce does the ball experience to accelerate from rest to 73 m/s?

Answers

Answer:

3.65 x mass

Explanation:

Given parameters:

Time = 20s

Initial velocity = 0m/s

Final velocity = 73m/s

Unknown:

Force the ball experience = ?

Solution:

To solve this problem, we apply the equation from newton's second law of motion:

F = m $(v - u)/(t)$

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

So;

F = m ( $(73 - 0)/(20)$ ) = 3.65 x mass

Final answer:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, use Newton's second law of motion.

Explanation:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).

Since the ball starts from rest, its initial velocity (vi) is 0 m/s. The final velocity (vf) is 73 m/s. The time (t) taken for the impact is given as 2 x 10 seconds. So, the acceleration (a) can be calculated using the formula a = (vf - vi) / t.

Substituting the given values into the equation, we have a = (73 - 0) / (2 x 10) = 3.65 m/s^2.

Now, we can find the force (F) using the formula F = m * a. If the mass of the ball is known, we can substitute it into the equation to find the force experienced by the ball.

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NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider such a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?