Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
Answer:
impulse = 8820 kg·
or 8820 N·s
Explanation:
Impulse J is equal to the average force 
multiplied by the elapsed time Δt or in equation form, J = Δt
As long as your force of 450 N is constant then that value is your average force and your elapsed time is 19.4 seconds.
Multiply these values.
You will get an impulse of 8820 kg· or 8820 N·s.
(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
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Answer:
Buffers
Explanation:
A buffer solution is a solution containing weak acids and their salts or weak bases and their salts.
A buffer solution is an equilibrium system that resists changes in pH or pOH when a small amount of an acid or base is added hence it is a solution of fairly constant pH value.
Answer: 10 and 35 degrees
Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment
