Two particles of masses m1m1m_1 and m2m2m_2 (m1

the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂

Explanation:

The center of mass, the point of a system where all external forces can be applied, is defined by

        = 1 /M ∑

where M is the total mass of the system, x_{i} m_{i} are the position and mass of each item in the system,

let's apply this equation to our case

the total mass is

    M = m₁ + m₂

for the calculations we must fix a reference system, we will place it in the second mass (m₂)

     x_{cm}= 1/M (m₁ d + m₂ 0)

where d is the distance between the two masses in this case d = 10 m

    x_{cm} = m₁ / (m₁ + m₂) d

    x_{cm} = m₁ / (m₁ + m₂) 10

as the mass m₁ <m₂

Let us analyze the answer if the masses sides were equal the center of mass would be x_{cm} = 5 m, but since m₁ <m₂ the center of mass must be closer to m₂.

Therefore the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas