Physics, 14.12.2019 00:31 layla07

A300-kg object and a 600-kg object are separated by 4.60 m.

(a) find the magnitude of the net gravitational force exerted by these objects on a 35.0-kg object placed midway between them. n
(b) at what position (other than an infinitely remote one) can the 35.0-kg object be placed so as to experience a net force of zero from the other two objects

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Answer from: Alex9089435028

$1.32391\times 10^{-7}\ N$

2.69 m

Explanation:

M = Mass in the middle = 35 kg

m_1 = 600 kg

m_2 = 300 kg

r = Distance

The net force between the objects would be

$F=\dfrac{GMm_1}{r^2}-\dfrac{GMm_2}{r^2}\\\Rightarrow F=\dfrac{GM(m_1-m_2)}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 35(600-300)}{2.3^2}\\\Rightarrow F=1.32391\times 10^{-7}\ N$

The magnitude of the net gravitational force exerted by these objects is $1.32391\times 10^{-7}\ N$

Now,

$\dfrac{GMm_1}{r^2}=\dfrac{GMm_2}{(4.6-r)^2}\\\Rightarrow \dfrac{m_1}{r^2}=\dfrac{m_2}{(4.6-r)^2}\\\Rightarrow r^2-9.2r+21.16=\dfrac{m_2}{m_1}r^2\\\Rightarrow r^2-9.2r+21.16=\dfrac{300}{600}r^2\\\Rightarrow r^2-9.2r+21.16=0.5r^2\\\Rightarrow 0.5r^2-9.2r+21.16=0\\\Rightarrow 50r^2-920r+2116=0$

Solving the above equation we get

$r=\frac{-\left(-920\right)+\sqrt{\left(-920\right)^2-4\cdot \:50\cdot \:2116}}{2\cdot \:50}, \frac{-\left(-920\right)-\sqrt{\left(-920\right)^2-4\cdot \:50\cdot \:2116}}{2\cdot \:50}\\\Rightarrow r=15.7, 2.69$

So, the distance at which the force will ben zero is 2.69 m

Answer from: sannai0415

Explanation:

m1 = 300 kg

m2 = 600 kg

d = 4.6 m

(a) m = 35 kg

Let the force between m1 and m is F1. Use the Newton's gravitation law

$F_{1}=\frac{Gm_{1}m}{\left (0.5d \right )^{2}}$

$F_{1}=\frac{6.67\times 10^{-11}\times 300\times 35}{2.3\times 2.3}$

F1 = 1.324 x 10^-7 N (towards m1, i.e., leftwards)

Let the force between m2 and m is F2. Use the Newton's gravitation law

$F_{1}=\frac{Gm_{2}m}{\left (0.5d \right )^{2}}$

$F_{1}=\frac{6.67\times 10^{-11}\times 600\times 35}{2.3\times 2.3}$

F1 = 2.65 x 10^-7 N (towards m2, i.e., rightwards)

Net force on m is

F = F2 - F1

F = (2.65 - 1.324) x 10^-7

F = 1.33 x 10^-7 N towards right

(b) Let it is placed at a distance r from m1 so that the net force is zero.

$\frac{Gm_{1}m}{r^{2}}=\frac{Gm_{2}m}{\left ( d-r \right )^{2}}$

$\frac{300}{r^{2}}=\frac{600}{\left ( d-r \right )^{2}}$

1.414 r = 4.6 - r

2.4 r = 4.6

r = 1.92 m

Thus, the net force on 35 kg is zero at a distance of 1.92 m from 300 kg.

A300-kg object and a 600-kg object are separated by 4.60 m. (a) find the magnitude of the net gravi

Answer from: Quest

elly's long hair:

a. if the salon only takes cash, then this isn't the problem. it doesn't matter how much she has in her checking account. she can't get a haircut until she has the cash.

b. if she has overdraft protection, then there's no problem. she can go ahead and write a check that her account can't cover, then sometime later, pay the bank some outrageous usury for loaning her a few bucks for a few days, but with beautiful hair.

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justin can't pay the full balance:

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i'd say that justin's best best option isn't on this list of choices. what he has to do: pay as much as he can at least the minimum payment plus as much more as he can and then avoid adding more debt.

Answer from: Quest

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.

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A300-kg object and a 600-kg object are separated by 4.60 m.

(a) find the magnitude of t...

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