Newton's Laws question, please help me:

Two objects of different masses accelerate at the same rate. According to Newtonâ€™s second law of motion, the more massive obje

babymother [125]

Newton's Second law of motion is

F = m * a ( force = mass * acceleration).

So if a is constant then a = F 1 / m1 and a = F2 / m2 then

F1 /m1 = F2 / m2 and if m1 > m2 then

F1 must be greater than F2.

Answer is C.

7 0

3 years ago

Hiii :( this is the second question and not making sense to me at all please help <3 it’s due soon

harkovskaia [24]

Answer:

Socratic app

Explanation:

it will help you

7 0

3 years ago

How do you know when motion has occurred?

Masteriza [31]

An object in motion changes position

7 0

4 years ago

The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;

$\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)$

Therefore;

$\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)$

From equation (4) we now make the new force $F_{o2}$ the subject of formula as follows;

${F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}$

$Q_o$ then cancels out from both side of the equation, hence we obtain the following;

$3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)$

From equation (4) we can now write the following;

$F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}$

This could also be expressed as follows;

$F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2$

3 0

4 years ago

When you lift a book from the ground to your desk, what kind of work do you do negative or positive by lifting the book what do

skelet666 [1.2K]

It would be kinetic energy. Let's say the book is weighs 10 Newtons you need to use a force of 10 Newtons to lift the book. In other words it's positive. As you move the book you're giving it energy. Namely potential energy which will turn to kinetic energy if you let it go. So you're changing it's position and energy.

7 0

4 years ago