Acceleration = Change in Velocity / time
a = (v - u) / t
Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²
A proper record of the changes in velocity with the corresponding time would help find the acceleration.
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
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Answer:
X=92.49 m
Explanation:
Given that
u= 21 m/s
h= 97 m
Time taken to cover vertical distance h
h= 1/2 g t²
By putting the values
97 = 1/2 x 10 x t² ( g = 10 m/s²)
t= 4.4 s
The horizontal distance
X= u .t
X= 21 x 4.4
X=92.49 m
Decreases density because gases have less density than liquids
The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
<h3>
Time of motion of the projectile</h3>
The time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
- h is height of the cliff
- v is velocity
- t is time of motion
265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
<h3>Final velocity of the projectile</h3>
vyf = vyi + gt
where;
- vyf is the final vertical velocity
- vyi is initial vertical velocity
vyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
- vxf is the final horizontal velocity
- vxi is the initial horizontal velocity
vxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
- vf is the speed of the projectile when it strikes the ground below
vf = √(149.322² + 130.8²)
vf = 198.51 m/s
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