A man is riding his 4-wheeler at 60 km/hr. If he is riding it constantly at

In reaching her destination, a backpacker walks with an average velocity of 1.34 m/s, due west. This average velocity results be

Ivanshal [37]

We can define average velocity as:

av = (total distance travelled)/(total time)

Only using the above equation, we will find that she walked 807.3m due east.

The given information is:

The average velocity is 1.34 m/s due west.

(we can define west as the positive side and east as the negative side).

We know that first, she hikes 6.44km due west with an average velocity of 2.68 m/s.

To get the total time it took we write the equation:

2.68 m/s = (6,440m)/(time)

time = 6,440m/(2.68 m/s) = 2,402.9 seconds.

Then she hikes a distance D due east with an average velocity of 0.447 m/s, because she goes due east, we will write -D in the equations.

Now using the same equation as before, we can get the time as:

time' = -D/(-0.447 m/s) = D/(0.447 m/s)

Now the equation for the total average velocity will be:

1.34 m/s = (6,440m - D)/(2,402.9 s + D/(0.447 m/s))

Now we need to solve this for D.

(1.34 m/s)*(2,402.9 s + D/(0.447 m/s)) = (6,440m - D)

(1.34 m/s)*(2,402.9 s) + (1.34 m/s)*(D/(0.447 m/s) =6,440m - D

(1.34 m/s)*(D/(0.447 m/s) + D = 6,440m - (1.34 m/s)*(2,402.9 s) = 3,229.1 m

D*(1 + (1.34m/s)/(0.447 m/s)) = 3,229.1 m

D*4 = 3,229.1 m

D = 3,229.1 m/4 = 807.3 m

We can conclude that she walked 807.3m due east.

If you want to learn more, you can read:

brainly.com/question/862972

8 0

3 years ago

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Derek runs 4 laps around the track. If each lap around the track is 0.25 miles long, and he starts and stops in the same locatio

Sergeeva-Olga [200]

1.00 1 mile 1 mile 1.00

5 0

3 years ago

Read 2 more answers

Can someone help me find the answer?

Elena L [17]

Wavelength, because of how it lengths out from the rest.

6 0

3 years ago

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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

For the course of upward acceleration:

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago

Will give brainliest! 50 points!!!

tatiyna

Answer:

C

Im not totally sure but that's what my science teacher taught me sorry if is wrong

6 0

3 years ago