Answer:
(a)20.65g
(b)0.19m
Explanation:
(a) The total mass would be it's mass per length multiplied by the total lenght
0.355(50 + 23*0.355) = 20.65 g
(b) The center of mass would be at point c where the mass on the left and on the right of c is the same
Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g



Answer:
The slower the intended velocity, the closer the force expressed comes to equalling the linear inertia of the load (i.e. the amount of force needed to hold the weight motionless). From Equation 1, force is inversely proportional to time
Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:
D = 1/2 a T²
Distance = (1/2)·(acceleration)·(time²)
This question gives us the acceleration and the distance, and we want to find the time.
(9,000 m) = (1/2) (20 m/s²) (time²)
(9,000 m) = (10 m/s²) (time²)
Divide each side by 10 m/s²:
(9,000 m) / (10 m/s²) = (time²)
900 s² = time²
Square root each side:
<em>T = 30 seconds</em>
Explanation:
Let us assume that forces acting at point B are as follows.

= 0 ...... (1)
= 0
= 0 .......... (2)
Hence, formula for allowable normal stress of cable is as follows.

T = 
= 3925 kip
From equation (1),
= -3925
= -3925
= 12877.29 kip
From equation (2), -12877.29 (Cos 60) + W = 0
= 0
W = 6438.64 kip
Thus, we can conclude that greatest weight of the crate is 6438.64 kip.
Answer:
El avión recorrió 45 km en los 180 s.
Explanation:
La relación entre velocidad, distancia y tiempo se da de la siguiente manera;

Por lo cual los parámetros dados son los siguientes;
Velocidad = 900 km/h = 250 m / s
Tiempo = 180 s
Estamos obligados a calcular la distancia recorrida
De la ecuación para la velocidad dada arriba, tenemos;
Distancia recorrida = Velocidad pf viaje × Tiempo de viaje
Distancia recorrida = 900 km/h × 180 s = 900
Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km
Por lo tanto, el avión viajó 45 km en 180 s.