Answer:
0.62 m/s² at 68° S of E
Explanation:
Net force north = 12 - 70 = -58 N
Net force east = 33 - 10 = 23 N
Net force = √(-58² + 23²) = 62.3939... N
acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²
θ = arctan(-58/23) = -68.3691... ≈ 68° S of E
Answer:

Explanation:
From the question we are told that:
Height 
Radius 
Height of water 
Gravity 
Density of water 
Generally the equation for Volume of water is mathematically given by


Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by

Substituting dv


Therefore




![W=3420.84*0.25[2401-65536]](https://tex.z-dn.net/?f=W%3D3420.84%2A0.25%5B2401-65536%5D)


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Answer:
A
Explanation:
Please see the attached picture for the full solution.
Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.
Explanation:
Let's say right is positive and left is negative.
F₁ = -150 N
F₂ = 220 N
Fnet = F₁ + F₂
Fnet = -150 N + 220 N
Fnet = 70 N
The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.
And since Fnet isn't 0, the force is unbalanced and the motion is changing.