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Montano1993 [528]
3 years ago
6

an airplane accelerates down a runway at 5.20 m/s^2 for 32.8s until it finally lifts off the ground. determine the distance trav

eled before takeoff
Physics
1 answer:
hammer [34]3 years ago
6 0

Answer:

7,272.7m

Explanation:

Our knowns:

a = 5.2m/s^2

t = 32.8s

V_{i} = 0 m/s

Our Unknowns:

Distance (x): ?

We can use displacement formula:

ΔX = V_{i} t + \frac{1}{2} a t^{2}

Plug in knowns:

X = 0 + 1/2 ((5.2)(32.8))^2

Answer:

ΔX = 7272.7m

(I beleive this should be right) let me know if it helped!

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Umnica [9.8K]

Unless acted upon by an external force, an object in motion tends to stay in motion, and an object at rest prefers to stay at rest.

We can also see that we regress when a car abruptly begins to move from a condition of rest. The inertia of repose also contributes to this. Our bodies have a tendency to stay in a condition of rest, but as soon as the automobile starts moving, we start to regress.

Unless influenced by an imbalanced force, an item at rest stays at rest, and an object in motion keeps moving in a straight path at a constant pace. Second Law of Motion by Newton.

Learn more about motion here-

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8 0
2 years ago
Four forces act at a point in the image above. calculate the resultant.
levacccp [35]

Answer:

50° and 60°

Explanation:

hope this help

3 0
3 years ago
ASAP When salt is dissolved in water, the result is a A. gas B. compound C. homogeneous mixture D. heterogeneous mixture
alisha [4.7K]
It’s actually C. homogeneous compound because when they combine you can’t separate them
4 0
3 years ago
Read 2 more answers
You have 50L of water, it froze. So find its new volume. Density of water is 1000kg/m^3 or 1kg/L, density of ice is 920kg/m^3 or
ad-work [718]

Answer:

D (density) = Mass / Volume

V (ice) = Mass / Density = 50000 g / .92 kg / L

V = 50 kg / .92 kg / L  

1 Liter of water weighs 1 kilogram = 50 L weighs 50000 g

V = 54.3 L     the mass does not change upon the change of phase (freezing)

4 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
4 years ago
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