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2 years ago

.An airplane accelerates down a runway at

Physics

1 answer:

solniwko [45]2 years ago

6 0

Starting from rest, the plane travels a distance

x = 1/2 at²

with acceleration a after time t. In this instance, it travels

x = 1/2 (3.20 m/s²) (32.8 s)²

x ≈ 1720 m

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Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10

nydimaria [60]

Answer:

Explanation:

The formula for this is

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

$7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}$ and moving things around to solve for r:

$r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }$ Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

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2 years ago

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

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2 years ago

What will occur when the trough of Wave A overlaps the trough of Wave B?

Pavel [41]

C because the wave length was longer

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3 years ago

Read 2 more answers

At the instant shown in the diagram, the car's centripetal acceleration is directed

serg [7]

Answer:

At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.

Explanation:

The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.

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2 years ago

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galben [10]

Answer:

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Explanation:

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3 years ago