Answer:
The speed of water must be expelled at 6.06 m/s
Explanation:
Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so
:

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for
:


The minus sign indicates the velocity of the water is opposite the velocity of the octopus.
Answer:
T =176 N
Explanation:
from diagram
F -(m_1+m_2_g) = (m_1+m_2_g)a
440 - (6+4)g = (6+4)a
a =\frac{440-10*9.8}{10}
a =34.2 m/s^2
frrom free body diagram of mass m2 = 4kg
T -m_2g =m_2a
T = m_2(g +a)
T = 4(9.81+34.2)
T =176 N
Answer:
v_y = v_{oy} - g t
where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses
Explanation:
In this exercise you are asked to observe the change in velocity in a projectile launch.
If we assume that the friction force is small, the velocity in the x-axis must be constant
vₓ = v₀ₓ
Therefore, the arrow (red) that represents this movement must not change in magnitude.
In the direction of the y axis, the acceleration of gravity is acting, so the magnitude of the velocity in this axis changes
v_y = v_{oy} - g t
where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses
Answer: The answer is D.
Explanation: Momentum can be defined as mass in motion. All objects have mass so if an object is moving, then it has momentum it has its mass in motion. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.