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3 years ago

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

Physics

1 answer:

ioda3 years ago

6 0

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

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sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

Average Acceleration

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that.

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

$\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}$

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$\boxed{a=9\ cm/s^2}$

6 0

3 years ago

How much force is required to pull a spring 3.0 cm from

avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

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4 0

4 years ago

A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes

storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2π (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

a = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

F = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

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3 years ago

A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)

Illusion [34]

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

$W=mg\Delta h$

where

m is the mass of the object

g is the acceleration of gravity

$\Delta h$ is the change in height of the object

For the object in this problem, we have:

m = 5 kg

$g=10 m/s^2$

$\Delta h = 2 m$

Substituting into the equation,

$W=(5)(10)(2)=100 J$

Learn more about work:

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3 years ago

What is the role of the nervous system?

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3 years ago