Answer:
Push and pull is the 3rd motion.
<span>Force F = 280 N
Angle with the ground = 40 degrees
Weight of the Lawnmower = 350 N
Velocity is constant so Acceleration is 0
So Forward force Ff = F cos theta = 280 cos40
Frictional force with resists to back Fb = (u x Force from pressure) + vertical component of Force, where u is the coefficient of friction.
Fb = (u x m x g) + (u x 280sin40)
AS Ff = Fb => 280 cos40 = u x ((m x g) + 280sin40)
u = 280 cos40 / ((350 x 9.81) + 280sin40) = 214.49 / () = 0.405
So the coefficient of friction u = 0.405</span>
Answer:
The work done on the air is 4.699 x 10⁶ Joules
Explanation:
Given;
increase in air volume, ΔV = 0.227 m³
net pressure of the air, P = 2.07 x 10⁷ Pa
The work done on the air is given by;
W = PΔV
Where;
W is the work done on the air
P is the net pressure
ΔV is the increase in air volume
Substitute the given values and solve for work done;
W = (2.07 x 10⁷ Pa) (0.227 m³)
W = 4.699 x 10⁶ Joules
Therefore, the work done on the air is 4.699 x 10⁶ Joules
Answer:
The sediment settled with the largest particles at the bottom and the smaller at the top.
-10/3m/s2. The formula for acceleration is (Final speed - initial speed)/ time.
Thus, (0m/s - 10m/s)/3s is 10m/s/3s which then can be simplified to /10/3m/s2