Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Respuesta:
Explicación:
Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!
Answer:
F=126339.5N
Explanation:
To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:
Fw=W+Fi+F
where
Fw= force or weight exerted by the water column above the submarine.
To calculate Fw, we can use:
Fw=h. γ. A
h=height
γ=
specific weight of seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w represents the hatch weight = 200N
Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:
Fi=PA=101325x0.7=70927.5N
Finally, the force needed to open the hatch is determined by the original equation:
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
Answer:
a) 
b) the distance the motorcycle covers is 155 m
Explanation:
Let
denote the variables. Next, we analyze the motion equation for the accelerating motorcycle alongside the constant speed of the car:

where:
represents the motorcycle's speed at time 2
is the steady velocity of the car
indicates the initial speeds of both vehicle types at time 1
d signifies the distance separating the car and motorcycle at the initial moment
x is the distance the car travels from time 1 to time 2
Solving the equations provides:
![\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dcar%26motorcycle%5C%5Cx%3Dv_0%5CDelta%7Bt%7D%26x%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%7D%29%20%5CDelta%7Bt%7D%5Cend%7Barray%7D%5Cright%5D)

For the second query, we determine x+d by applying the car’s motion equation to compute x:
