The answer is A. speed
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Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²
Answer:
The velocity of the skateboard is 0.774 m/s.
Explanation:
Given that,
The spring constant of the spring, k = 3086 N/m
The spring is stretched 18 cm or 0.18 m
Mass of the student, m = 100 kg
Potential energy of the spring, 
To find,
The velocity of the car.
Solution,
It is a case of conservation of energy. The total energy of the system remains conserved. So,
v = 0.774 m/s
So, the velocity of the skateboard is 0.774 m/s.
Answer:
speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Explanation:
Given:
mass of truck M = 1370 kg
speed of truck = 12.0 m/s
mass of car m = 593 kg
collision is elastic therefore,
Applying law of momentum conservation we have
momentum before collision = momentum after collision
1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)
Also for a collision to be elastic,
velocity of approach = velocity of separation
12 -0 = v2-v1 ....(ii)
using (i) and (ii) we have
So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
