What constant acceleration is required to increase the speed of a car from 27 mi/h to 53 mi/h in 3s?
1 answer:
You might be interested in
Refer to the diagram shown below.
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s
Answer:
Explanation:
(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)
Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:
Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:
Ahora se reemplaza en la ecuación de período:
La aceleración experimentada por el planeta es:
Se reemplaza en la ecuación de período:
La distancia del planeta con respecto al sol es finalmente despejada:
Finalmente, se sustituyen las variables y se determina la distancia: