How do butterfly dest on there wings

according to newtons second law of motion, when an object is acted on by an unbalanced force, how will that object respond?

olga_2 [115]

When an object is acted on by an unbalanced force, then that object will accelerate.

6 0

2 years ago

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A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2

ZanzabumX [31]

The magnitude of gravity is expressed in terms of its acceleration. So the magnitude of ' g ' at that altitude is exactly 6.5 m/s^2.

8 0

3 years ago

A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow

ra1l [238]

Answer:

No.

Explanation:

The "guide to Engineering and land surveying" for professional engineers and land surveyors by the California board reviews that an unlicensed person cannot be a sole owner of an engineering business, unless there is partnership with a licensed engineer.

4 0

3 years ago

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A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (

lions [1.4K]

initially, the car is traveling at 5.0 m/s.

so, we know acceleration for changing velocity is :

a = (v-v $_{o}$ )/t ..........(i)

where v is the final velocity

v $_{o}$ is the initial velocity

t is the time taken to change velocity

Now, as per the question :

initial velocity, v $_{o}$ =5.0 m/s

final velocity, v =11 m/s

time taken, t = 3 s

putting the values in equation (i),

a = ( 11-5 )/3

a = 2 m/s²

Therefore, a, after 3 s, is <em>2 m/s².</em>

4 0

6 months ago