Physics, 25.12.2019 19:31 jstamey757

(ii) two objects attract each other gravitationally with a force of 2.5 × 10 ^–10 n when they are 0.25 m apart. their total mass is 4.00 kg. find their individual masses.

Answers: 2

Show answers

Answers

Answer from: Zaxx2974

The individual masses of the two objects are 0,06 kg and 3,94 kg.

Explanation:

To solve this we need to use Newton's law of universal gravitation that states that the force of gravity between two massive bodies 1 and 2, is proportional to their masses M₁ and M₂ and inversely proportional to the square of the distance d that separates them:

$F=\frac{G M_{1}M_{2}}{d^{2} }$

with $G=6,67 x 10^{-11} \frac{Nm^{2} }{kg^{2} }$ being the universal gravitational constant.

We are told:

the value of F when it says that the objects attract each other gravitationally with a force of $2.5x10^{-10}N$ ,the value of d when it says that they are 0.25 m apart.

When it says that their total mass is 4.00 kg it means that:

$M_{1} +M_{2} =4.00\ kg$

$M_{2} =4.00\ kg-M_{1}$

replacing all of this in the original equation $F=\frac{G M_{1}M_{2}}{d^{2} }$ we get

$2.5x10^{-10}N=6.67x10^{-11}\frac{Nm^{2}}{kg^{2}} \frac{M_{1}(4.00kg-M_{1})}{(0.25)^{2}m^{2}}$

$2.5x10^{-10}N=\frac{6.67x10^{-11}Nm^{2}}{(0.25)^{2}m^{2}kg^{2}}(M_{1}4.00kg-M_{1} ^{2})$

$2.5x10^{-10}N=\frac{6.67x10^{-11}N}{(0.25)^{2}kg^{2}}(M_{1}4.00kg-M_{1} ^{2})$

$\frac{2.5x10^{-10} }{6.67x10^{-11}}(0.25)^{2}kg^{2}=4.00kg\ M_{1}-M_{1} ^{2}$

if we put all terms in one side we get

$M_{1} ^{2} -4.00kg\ M_{1} +\frac{2.5x10^{-10}}{6.67x10^{-11}}(0.25)^{2}kg^{2} =0$

$M_{1} ^{2} -4.00kg\ M_{1} +0.23\ kg^{2} =0$

we have to solve a quadratic equation $ax^{2} +bx+c=0$ where a=1, b=-4.00 and c=0.23. Here x=M₁.

We replace these values in the quadratic formula for the roots:

$x=\frac{-b+-\sqrt{b^{2}-4.a.c}}{2a}$

so we get

$x=\frac{4.00+-\sqrt{16-4.1.0.23}}{2}$

$x=\frac{4.00+-\sqrt{16-0.92}}{2}$

$x=\frac{4.00+-\sqrt{15.08}}{2}$

$x=\frac{4.00+-3.88}{2}$

the possible roots are

$x_{1} =\frac{4.00+3.88}{2}=\frac{7.88}{2}=3.94$

$x_{2} =\frac{4.00-3.88}{2}=\frac{0.12}{2}=0.06$

given that M₁+M₂=4.00 kg the possible values of M₂ are then

$M_{2}=4.00kg-3.94kg=0.06kg$

$M_{2}=4.00kg-0.06kg=3.94kg$

Either one of the individual masses M₁ and M₂ could be 0.06 kg or 3.94 kg.

Answer from: Quest

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation:

Answer from: Quest

the answer would be a

Another question on Physics

Physics, 22.06.2019 14:10

The number of passengers who arrive at the platform of a subway station for the 10 am train is a random variable with a mean of 120 and a variance of 16. find the lower bound of the probability that there will be between 100 and 140 passengers (round off to second decimal place).

Answers: 3

Answer

Physics, 22.06.2019 22:00

If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?

Answers: 1

Answer

Physics, 22.06.2019 23:00

During the past 150 years in the united states, how did energy consumption compare to population growth

Answers: 1

Answer

Physics, 23.06.2019 01:30

Match each type of lever with the correct diagram. 1. first-class 2. third-class 3. second-class