The train would need the greatest amount of force due to weight! If you think of it, a baseball won't need much force to stop it, but if you have a heavy train, it will need excessive force to stop the train. The answer would be #3
I hope this answer helps!
Sorry if it doesn't make sense, as I don't know that much about physics! I am just thinking of what makes sense.
Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
![a = \frac{v}{t}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bv%7D%7Bt%7D%20)
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
![a = \frac{49}{0.1} \\ a = \frac{490}{1} \\ a = 490](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B49%7D%7B0.1%7D%20%20%5C%5C%20a%20%3D%20%20%5Cfrac%7B490%7D%7B1%7D%20%20%5C%5C%20a%20%3D%20490)
The acceleration is 490 m/s^2
Answer:
1.73 m/s²
3.0 cm
Explanation:
Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.
Sum of forces in the y direction:
∑F = ma
T cos 10° − mg = 0
T cos 10° = mg
T = mg / cos 10°
Sum of forces in the x direction:
∑F = ma
T sin 10° = ma
mg tan 10° = ma
g tan 10° = a
a = 1.73 m/s²
Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
D − mg = 0
D = mg
½ ρₐ v² C A = ρᵢ V g
½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g
3 ρₐ v² C = 8 ρᵢ r g
r = 3 ρₐ v² C / (8 ρᵢ g)
r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))
r = 0.030 m
r = 3.0 cm
Answer:
professional communication style is expected to be formal while the casual is informal.
Explanation: