A spring toy jumps up from the floor and comes back down to the floor. Its initial speed is 12 m/s. What is the
1 answer:
Answer:
.
Explanation:
In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.
In this example, the spring toy is the projectile. Refer to the diagram attached.
- The spring toy started with a velocity of
, meaning that its initial KE is non-zero.
- On the way up, the KE of this spring toy gets converted to GPE.
- At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
- As the toy returns to the ground, the GPE of this toy gets converted back to KE.
- The GPE of this toy is zero when the toy is on the ground.
In other words, the kinetic energy (KE) of this toy would be when it is at the top of the trajectory and GPE is maximized.
Since the KE of this toy at the top of the trajectory would be , the velocity of this toy at that moment would also be
.
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Answer:
(a) The force between them quadruples
Explanation:
According to coulomb's law, initial force between the two charged objects is given as;
where;
k is coulomb's constant
q₁ is the charge on the first object
q₂ is the charge on the second object
r is the distance between the two objects
When the charges on both objects are doubled, then;
q₁ = 2q₁
q₂ = 2q₂
Force between the two charged objects will become
Therefore, the force between them quadruples
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
