Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t = 
t = 
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
Answer:
138,516,546.9 horas.
Explanation:
Tenemos que usar la ecuación:
Velocidad = distancia/tiempo
Acá tenemos:
Velocidad = 0.3m/s
distancia = 149597870700 m
y queremos resolver la ecuación para el tiempo:
0.3m/s = 149597870700m/tiempo.
tiempo = 149597870700m/(0.3m/s) = 498,659,569,000 s
y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:
498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.
The precipitate is the barium sulfate, or 3) BaSO4. This is because it is a solid (as seen by the (s)), unlike the other aqueous product. The fact that it is a solid means that it is insoluble in water and therefore a precipitate.
Hope this helps!
Answer:
Correct answer: t = 2.86 seconds
Explanation:
We first use this formula
V² - V₀² = 2 a d
where V is the final velocity (speed), V₀ the initial velocity (speed),
a the acceleration and d the distance.
We will calculate the acceleration from this formula
a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10
a = 0.525 m/s²
then we use this formula
V = V₀ + a t => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds
t = 2.86 seconds
God is with you!!!
Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
