Answer:
Aceleración, a = 2 m/s²
Explanation:
Dados los siguientes datos;
Velocidad inicial = 108 km/h
Tiempo = 10 segundos
Velocidad final = 36 km/h
To find the average acceleration;
Conversión:
36 km/h to meters per seconds = 36*1000/3600 = 10 m/s
108 km/h to meters per seconds = 108*1000/3600 = 30 m/s
I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;
Dónde;
V es la velocidad final.
U es la velocidad inicial.
a es la aceleración.
t es el tiempo medido en segundos.
Sustituyendo en la fórmula, tenemos;
Aceleración, a = 2 m/s²
Answer:
First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters
Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - 
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
Answer:
Explanation:
Let the volume of the unknown bulb = X L
The volume of the system , after opening valve = (X + 0.72 L )
Use Boyles law gas equation,
P1V1 = P2V2 ( at temperature is constant )
Given:
P1 = 1.2 atm
P2 = 683 torr
Converting mmHg to atm,
1 atm = 760 mmHg(torr)
683 mmHg = 683/760
= 0.8987 atm
1.2X = 0.8987*(X + 0.720)
1.2X = 0.8987X + 0.6471
0.3013X = 0.6471
X = 2.15 L
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb