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3 years ago

A space ship to the moon covered the 216,000 km in 72 hours. What was it’s average velocity

Physics

2 answers:

Dahasolnce [82]3 years ago

8 0

Answer:3000km/h

Explanation:

speed=distance/time

216,000km/72hrs

=3000km/h

goldenfox [79]3 years ago

6 0

Well, first of all, if there are astronauts aboard that ship, I hope they don't try to climb out yet. If they pointed straight at the moon and covered 216,000 km so far, they still have another 168,000 km to go before they get there.

Average Speed = (distance) / (time)

Speed = (216,000 km) / (72 hours)

<em>Average Speed = 3,000 km/hr</em>

also

Speed = (216,000 km) x (1000m/km) / (72 hr x 3600sec/hr)

<em>Average Speed = 833.33 m/s</em>

There's not enough information given to find the ship's velocity.

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

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3 0

2 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

The ratio of the output work to input force is the _ of the machine ?

Colt1911 [192]

<h3><u>Answer;</u></h3>

Mechanical advantage

<h3><u>Explanation;</u></h3>

Mechanical advantage is the ratio of force output from a machine divided by the force input into the machine.
Mechanical advantage measures the machine's force-magnifying effect. It is an advantage gained by using simple machines to accomplish work with less effort.
The formula is; M.A = output force/ Input force

3 0

3 years ago

Read 2 more answers

An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De

gayaneshka [121]

We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
$F=T-W$
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
$a= \frac{F}{m}$
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
$F=T-W=6000 N-5000 N=1000 N$
To find m, we can use the weight of the system. In fact, the weight of an object is given by
$W=mg$
where $g=9.81 m/s^2$ . Solving for m, and using W=5000 N, we find
$m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg$

and at this point, we can calculate the acceleration of the system (elevator+people):
$a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2$
and the acceleration has the same direction of the resultant force, so upward.

8 0

3 years ago

A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.30 m from the slits. If the bright

kumpel [21]

Answer:

λ = 596 nm.

Explanation:

Fringe width = λ D / d

λ is wave length , D is screen distance and d is slit separation.

Putting the values

1.62 x 10⁻² =( λ x 5.3 ) / .195 x 10⁻³

$\lambda=\frac{1.62\times10^{-2}\times195\times10^{-6}}{5.3}$

λ = 596 nm.

8 0

3 years ago