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crimeas [40]
3 years ago
11

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

^- 8 N how far apart are centers of mass?

Physics
1 answer:
nydimaria [60]3 years ago
8 0

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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The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d
r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let F_1\ and\ F_2 are two forces. The resultant of two forces acting on the same point is given by :

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}

Where \theta is the angle between two forces

When \theta=0 i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}

When \theta=90^{\circ} i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}

F_R=\sqrt{F_1^2+F_2^2}

When \theta=180^{\circ} i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}

F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

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3 years ago
How do oxygen and beryllium atoms transform into oxygen ion O2- and Be2 beryllium ion Be2? tysm
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On the whole, the metals burn in oxygen to form a simple metal oxide. Beryllium is reluctant to burn unless it is in the form of dust or powder. Beryllium has a very strong (but very thin) layer of beryllium oxide on its surface, and this prevents any new oxygen getting at the underlying beryllium to react with it.
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three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a
valentinak56 [21]

Answer:

q_1=+0.375\ {10}^{-9}

Explanation:

Electrostatic Forces

The force exerted between two point charges q_1 and q_2 separated a distance d is given by Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{d^2}

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

\displaystyle q_3=+5\ 10^{-9}\ c

\displaystyle q_2=-3\ 10^{-9}\ c

The distance between q_3 and q_2 is

\displaystyle d_2=4cm=0.04\ m

The distance between q_3 and q_1 is

\displaystyle d_1=2cm=0.02\ m

We must find the value of q_1 such that

\displaystyle |F_3|=0

Applying Coulomb's formula for q_1 is

\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}

Now for q_2

\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}

If the total force on q_3 is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

\displaystyle F_{13}=F_{23}

\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}

Simplfying and solving for q_1

\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}

\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}

\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}

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Answer: B

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