Answer:
α = - 1.883 rev/min²
Explanation:
Given
ωin = 113 rev/min
ωfin = 0 rev/min
t = 1.0 h = 60 min
α = ?
we can use the following equation
ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t
⇒ α = (0 rev/min - 113 rev/min) / (60 min)
⇒ α = - 1.883 rev/min²
To change from mass to weight is Fw = 30 kg * 9.8 m/s^2 = 294 N. To change from weight to mass divide by gravity (9.8 m/s^2).
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
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Please check the attached picture
Answer:
the final velocity of the car is 59.33 m/s [N]
Explanation:
Given;
acceleration of the car, a = 13 m/s²
initial velocity of the car, u = 120 km/h = 33.33 m/s
duration of the car motion, t = 2 s
The final velocity of the car in the same direction is calculated as follows;
v = u + at
where;
v is the final velocity of the car
v = 33.33 + (13 x 2)
v = 59.33 m/s [N]
Therefore, the final velocity of the car is 59.33 m/s [N]
