What is the actual depth of rainfall shown in this rain gauge?
2 answers:
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Answer:
The vertical distance that the ski jumper fell is 417.45 m.
Explanation:
Given;
initial horizontal velocity of the jumper, = 26 m/s
horizontal distance of the jumper, dx = 240 m
The time of the motion is given by;
dx = Vₓt
t = dx / Vₓ
t = 240 / 26
t = 9.23 s
The vertical distance traveled by the diver is given by;
initial vertical velocity, , = 0
Therefore, the vertical distance that the ski jumper fell is 417.45 m.
The equilibrant force of the two given forces is 14.14 N.
<h3 /><h3 /><h3>What is equilibrant force?</h3>- This is a single force that balances other given forces.
The given parameters:
- First force, F₁ = 10 N
- Second force, F₂ = 10 N
- Angle between the forces, θ = 90⁰
The equilibrant force of the two given forces is calculated as follows;
Thus, the equilibrant force of the two given forces is 14.14 N.
Learn more about equilibrant force here: brainly.com/question/8045102
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
