What is the momentum of a 2000 kg mass moving at 3 meters per second
1 answer:
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In case 1, the torque is given by the product between the force and the arm:
In case 2, the torque is given by the product between the component of the force perpendicular to the arm and the arm itself, so we have:
and since
is larger than 1, than the torque in case 2 is larger than the torque in case 1.
In case 2, the torque is given by the product between the component of the force perpendicular to the arm and the arm itself, so we have:
and since
Answer:
The force applied on the big piston is 1306.67 N
Explanation:
Given;
force applied on small piston, F₁ = 200 N
diameter of the small piston, d₁ = 4.37 cm
radius of the small piston, r₁ = d₁/2 = 2.185 cm
Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²
Area of the big piston, A₂ = 98 cm²
The pressure of the piston is given by;
Where;
F₂ is the force on big piston
Therefore, the force applied on the big piston is 1306.67 N
Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:
which in our case becomes;
and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N