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3 years ago

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IrinaK [193]3 years ago

5 0

Answer:

dssdfsdsdfsf

Explanation:

sdsdfsdfsdf

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The average distance from Earth to the Moon is 384,000 km. In the late 1960s, astronauts reached the Moon in about 3 days. How f

Anon25 [30]

Answer:

They must have been traveling at 5333.33 km/h to cover that distance in 3 days.

That speed are 6,66 times higher than the speed of an aircraft jet.

Explanation:

d= 384000 km

t= 3 days = 3*24hr = 72hr

V= 384000km/72hr

V= 5333.33 km/h

comparison:

V1/V2= 5333.33/800

V1/V2= 6.66

3 0

3 years ago

Waves carry ______ from one place to another?<br><br> energy<br> crests<br> troughts<br> wavelengths

Leni [432]

Answer:

Energy.

Explanation:

Waves can travel through medium and can be without medium.

In electromagnetic waves the energy travels through vibration of electric field and magnetic field while in sound waves energy is transferred through vibration of particles through which sound waves are travelling or though vibration of air particles.

7 0

3 years ago

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A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

Question 3 You see the coordinates 5 E longitude, 10 N latitude. You do not need to look at a map in order to deduce that this l

san4es73 [151]

Answer:

a. Near both the equator and the prime meridian.

Explanation:

The equator is at 0 degrees latitude and the prime meridian is 0 degrees longitude.

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3 years ago

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