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3 years ago

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Why do we use the two-body problem to solve interplanetary trajectories, instead of including all of the appropriate gravitation

al forces that actually apply?

1 answer:

egoroff_w [7]3 years ago

6 0

Answer:

ur mom

hsheu7shrbrjxbfbbrndnifidfjf

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A box experiences a force of 2 N to the left and 3 N to the right. Which is true of the box's motion?

Nostrana [21]

Below are the choices:

<span> A) The box will slow down.
B) The box's velocity will be 1 m/s.
C) The box's velocity will not change.
D) The box will experience acceleration
</span>
The answer is D) The box will experience acceleration

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5 0

3 years ago

Read 2 more answers

La altura de un tornillo de banco respecto a la superficie es de 80 cm expresar dicha medida en pies..

Andrej [43]

Answer:

this measurement if feet is: 2.624672 ft

Explanation:

Notice that 80 cm can be expressed as 0.8 meters, and In order to convert from meters to feet, one needs to multiply the meter measurement times 3.28084. Therefore:

0.80 m can be written in feet as: 0.80 * 3.28084 feet = 2.624672 feet

3 0

3 years ago

When two ions form a bond, the overall charge of that compound will ALWAYS become..

Maslowich

Answer:

C. Neutral

Explanation:

Ions will combine in a way that the overall ionic compound will always be neutral.

8 0

3 years ago

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

2 years ago

What does the formula "F=m xa" mean?

strojnjashka [21]

Answer:

Force = Mass x Acceleration

Explanation:

5 0

3 years ago