Answer:
C) 50 m/s
Explanation:
With the given information we can calculate the acceleration using the force and mass of the box.
Newton's 2nd Law: F = ma
- 5 N = 1 kg * a
- a = 5 m/s²
List out known variables:
- v₀ = 0 m/s
- a = 5 m/s²
- v = ?
- Δx = 250 m
Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:
Substitute known values into the equation and solve for v.
- v² = (0)² + 2(5)(250)
- v² = 2500
- v = 50 m/s
The final velocity of the box is C) 50 m/s.
Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
Therefore, the relationship between force and motion is that it takes force to change the speed or direction of any object in motion.
The answer is A.number of protons in the nucleus.
Answer:
a. 1100 meters.
b. Between B and C
c.1. Between point D and E
c2. Between point D and E
d. 3.7 m/s.
Explanation:
The girl travels the distance of 1100 meters from starting to the end. There is no motion occurs between B and C due to no change of distance value from 200 meters. Between point D and E, the girls covers 500 meters long distance and also covers fastest distance between point D and E because between point D and E, the girl covers 500 meters distance in 30 seconds which is the highest of all. The average speed of the girls is 3.7 meter/seconds if we divide total distance i.e. 1100 meters by time which is 300 seconds.
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
