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3 years ago

How should you approach a dock when the wind or current is pushing you away from the dock?

Physics

1 answer:

erastova [34]3 years ago

8 0

Answer:

If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

MAVERICK [17]

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

now, equating both the distances

$23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

$t_2^2 = 15.143$

t₂ = 3.89 s

time taken by the motorcycle to catch the car is equal to 3.89 s

4 0

2 years ago

An electromagnetic wave of wavelength

Ivanshal [37]

Answer:

$4.01\cdot 10^{-7} m$

Explanation:

When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.

In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength in a vacuum (air is a good approximation of vacuum)

n is the refractive index of the medium

In this problem:

$\lambda_0 = 5.89\cdot 10^{-7} m$ is the original wavelength of the wave

n = 1.47 is the index of refraction of corn oil

Therefore, the wavelength of the electromagnetic wave in corn oil is:

$\lambda=\frac{5.89\cdot 10^{-7}}{1.47}=4.01\cdot 10^{-7} m$

8 0

2 years ago

Design an experimental set up to show that white light is made up of different colours of light

DaniilM [7]

Answer:

Issac Newton was the first to use a glass prism to obtain the spectrum of sunlight. He tried to split the colours of the spectrum of white light further by using another similar prism. However, he could not get any more colours. He then placed a second identical prism in an inverted position with respect to the first as shown.

This allowed all the colours of the spectrum to pass through the second prism. He found a beam of white light emerging from the other side of the second prism. This observation gave Newton the idea that the sunlight is made-up of seven colours.

Explanation:

6 0

2 years ago

. A weather balloon is inflated to a volume of 2.2 x 103 L with 37.4 g of helium. What is the density of helium in grams per lit

Margarita [4]

That is the mst best eway to find its solution.
37.4/2.2*10^3 = 0.017 gm/liter or 1.7*10^-2
so we conclude that option b is sorrect

6 0

3 years ago

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

The number of fringe is z = 3 fringes

The ratio is $I = 0.2545I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

$I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

$I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

6 0

3 years ago