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LenaWriter [7]
3 years ago
6

An airplane accelerates from rest down a runway at 3.20 m/s2 for 32.8 seconds until it lifts off the ground. Determine the dista

nce traveled and velocity before takeoff.
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

s=1721.344m  ,v=104.96m/s.

Explanation:

using thr equation of motion;

s=ut+\frac{1}{2}at^{2}

u=0, plane starts from rest,

s=\frac{1}{2}at^{2}

a=3.2m/s^{2}, t=32.8s \\ s=\frac{1}{2}*3.2*32.8^{2}

s=1721.344m

v=u+at

v=0 +3.2*32.8

v=104.96m/s

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John, paul, and george are standing in a strawberry field. paul is 14.0 m due west of john. george is 36.0 m from paul, in a dir
sdas [7]

Position of paul with respect to john is given as

14 m due west of john

r_{pj} = r_p - r_j = -14\hat i

position of George with respect to Paul is given as 36 m in direction 37 degree south of east

r_{GP} = r_G - r_p = 36cos37\hat i - 36 sin37\hat j

now we need to find the position of George with respect to John

r_{GJ} = r_G - r_j[\tex]now for the above equation we can add the two equations[tex]r_{Gj} = -14\hat i + 36 cos37\hat i - 36sin37\hat j

r_{Gj} = 14.75\hat i - 21.67\hat j

so the magnitude is given as

r = \sqrt{14.75^2 + 21.67^2} = 26.2 m

and direction is given as

\theta = tan^{-1}\frac{21.67}{14.75}= 55.75 degree

<em>so it is 26.2 m at an angle 55.75 degree South of east</em>

7 0
3 years ago
Read 2 more answers
A scientist in Northern California is studying the tree rings of a very old redwood tree. He notices that the oldest tree rings
Studentka2010 [4]

Answer: colder and drier

Explanation:

Trees don’t grow well in wet climates, nor when they are cold. The wider the rings, the healthier the tree is. Since it was narrow and then wide, was bad climate and then good.

5 0
3 years ago
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At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig
Anettt [7]

Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

<u>Second step:</u> we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;  

y = ut - \frac{gt^2}{2}

y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}

y_{1} = -19.6m

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1}  = 5.4m[/tex]

<u>Third step</u>: we determine the y-coordinate of ball 2 after 2 seconds

y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}

y_{2} = 10.4m

<u>Fourth step: </u>we determine the y-component of the center mass of the two balls

y = \frac{m_{1}y_{1} +m_{2}y_{2}}{m_{1} +m_{2} }

y = \frac{(0.5)X(5.4) +(0.25) X (10.4)}{(0.5 +0.25) }

y = 7.1m

<u>Fifth step:</u> we solve B part of the question; velocity of the center mass of the two balls

Velocity = \frac{distance of center mass of the two balls (y)}{time}

velocity = \frac{7.1 m}{2 s}

velocity = 3.55m/s

<u>Sixth step:</u> we solve C part of the question; acceleration of the center mass of the two balls

acceleration = \frac{velocity}{time}

acceleration = \frac{3.55}{2}

acceleration = 1.775 m/s^2

7 0
3 years ago
A post is wrapped two full turns around with a belt. The tension in the belt is 7500 N by exerting a force of 150 N on its free
ira [324]

Answer:

Coefficient of friction is

Ū = 0.31

Explanation:

T2 = T1* e^(ūơ)

Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,

ū = coefficient of friction

Ơ = 2pai * N

Where N = number of turns = 2

Ơ = 4pai

7500 = 150e^(ū*4pai)

50 = e^(ū *4pai)

lin 50 = 4pai * ū

Ū = 3.91/4pai

Ū = 0.31

4 0
3 years ago
How is mass related to momentum???
tresset_1 [31]

the formula for (momentum) P = (force) F * (Mass) M

P = FM

if mass = 0 momentum = 0

Hope this helps...

3 0
3 years ago
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