Answer:
V = 48.4m/s, θ = 73.6° below the horizontal.
Explanation:
Given h = 109.53m, x = range = 65m,
θ = 0°
This problem involves the concepts of projectile motion.
Let yo = h = 109.53m
y = final position on the y axis. = 0m (ground level)
y = yo + vosinθ – 1/2gt²
0 = 109.53 + vosin0° – 1/2×9.8×t²
0 = 109.53 – 4.9t²
4.9t² = 109.53
t² = 109.53/4.9 = 22.35
t = √22.35 = 4.73s
So it takes the ball t = 4.73 seconds to get to the ground from the launch point.
x = (vocosθ)×t
65 = (vocos0°)×4.73
65/4.73 = vo
Vo = 13.7m/s
Vx = Vox = Vocosθ = 13.7cos0° = 13.7m/s
Vy = Voy – gt = Vosinθ – gt
Vy = 13.7sin0° – 9.8×4.73 = –46.4m/s
V = √(Vy² + Vx²) = √(-46.4² + 13.7²)
V = 48.4m/s
θ = Tan-¹(vy/vx) = tan-¹(-46.4/13.7) = -73.6°
θ = 73.6° below the horizontal
V = 48.4m/s
Answer:
Explanation:Required formula is
W=F *S
W=work =?
F=force =5N
S=displacement =0.5m
please feel free to ask if you have any questions
Answer:
Option C: The solution begins to turn blue
Explanation:
When a copper wire is placed in a beaker containing a solution of silver nitrate(AgNO3), the copper (Cu) will reduce the positive silver ions (Ag+) to metallic silver (Ag). In this same process, Copper (Cu) is oxidized to produce Copper II ions (Cu2+).
The reaction will continue to progress and silver (Ag) crystals will begin to form on the Copper (Cu) wire and thus, we will observe that the solution becomes blue as a result of the formation of copper II (Cu2+) ions.
This is a single replacement reaction.
The weight of the automobile is 17,533.6 N.
<h3>
Weight of the automobile</h3>
The weight of the automobile is calculated as follows;
P = F / A
F = (W/4)
P = (W/4) / A
P = W/4A
W = 4AP
where;
W = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)
W = 17,533.6 N
Thus, the weight of the automobile is 17,533.6 N.
Learn more about weight here: brainly.com/question/2337612
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