a model rocket climbs 200 m in 4 seconds. if was moving 10 m/s to begin with, what is it’s final velocity?

1 answer:

8 0

So first Identify all the given Varibales so u can choose which Eqauton to use

D=200m
T=4s
Vi=10m/s
Vf=?

You should this equation

D= 0.50(Vf+Vi)T

Plug in the values

200= 0.50 (Vf+10) 4

Divide the 4 out of the right side and if you do sumthing to one side you gotta do it to the other

200 divided by 4= 0.50(Vf+10)
50= 0.50(Vf+10)

Now expand the 0.50

So 50= 0.5Vf + 5 (because 0.5 times 10 is 5)

Now get rid of the 5

50-5= 0.5Vf

45 =0.5Vf now Divide the 0.5 out

45 divided by 0.5 = Vf

And 45/0.5 is 90

So 90=Vf

Therefore the final Velocity is 90m/s

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uranmaximum [27]

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6 0

3 years ago

A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r

katen-ka-za [31]

Answer:

k = 167.33 N/m

Explanation:

The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
v = c = λ*f

where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =

4.92*10⁷ m.

Solving for f, we get the frequency of the radio waves:

f = 6.1 Hz

Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:

$\omega_{o}^{2} =\frac{k}{m} (1)$

Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

$\omega = 2*\pi *f (2)$

We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
$k = 4*\pi ^{2}*f^{2} *m = 4*\pi ^{2} * (6.1Hz)^{2} * 0.114 kg = 167.33 N/m$