0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)
A. Angular momentum is always conserved would be the correct answer.
This is because like linear momentum (mvmv), angular momentum (r×mvr×mv) is a conserved quantity, where rr is the vector from the center of rotation. For a skater holding a static pose, for each particle making up her body, the contribution in magnitude to the total angular momentum is given by mirivimirivi. Thus bringing in her arms reduces riri for those particles. In order to conserve angular momentum, there is then an increase in the angular velocity.
hope this helps!
C. Is the da answer to your question!
Answer:
quantity A is mass and quantity B is wright
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
