For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system
as we know that
now from above equation we have
so the speed of combined system is 2 m/s
The acceleration of the boxes depends on the mass and weight.
we have a mass of 7 and 8 kilograms
if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2.
It will leave you with 100 N.
finally take the sq rt of 100 to get 10
Answer:
<em>600N(downwards)</em>
Explanations
<em>600N(downwards)</em>
Mas of the person = 60kg
Acceleration due to gravity = -10m/s²
To get the earths pull on the person, we will use the Newton second law of motion;
Force = mass * acceleration;
Force = 60 * -10
Force- -600N
<em>Hence the earth gravitational pull on the person is 600N(downwards). It is downwards due to the negative sign.</em>
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