<span>Let m1=10kg and m2=5kg and for our calculations assume right is positive and up is positive (note: for block hanging, the x axis is vertical so tilt your head to help)
For m1
Sigma Fx = ma
T - m1gsin35 = m1a where T = tension
For m2
m2g - T = m2a
Add equation together
m1a + m2a = T-m1gsin35 + m2g - T
a(m1 + m2) = m2g - m1gsin35
a= (5*9.8 - 10*9.8*sin35)/(10 + 5)
a= -0.48m/s/s
So the system is moving in the opposite direction of our set coordinate system where we said right positive, its negative so its moving left therefore down the ramp</span>
In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object
according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is
total momentum before collision = total momentum after collision
P₁ + P₂ = P'₁ + P'₂
where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.
P'₁ - P₁ = - (P'₂ - P₂)
so clearly gain in momentum of one object is same as the loss of momentum of other object
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
Hope i helped
Have a good day :)
Answer:
E. downward and constant
Explanation:
Freefall is a special case of motion with constant acceleration because the acceleration due to gravity is always constant and downward. This is true even when an object is thrown upward or has zero velocity.
For example, when a ball is thrown up in the air, the ball's velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration ggg, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.