The answer is 12.5 kg because 250N / 20m/s^2
I hope that helped
The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.
Answer:
D. At x=0, it's acceleration is at a maximum
Explanation:
As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.
Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.
From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.
The same can be said as the box travels backward from point A to -A
Answer:
6.5e-4 m
Explanation:
We need to solve this question using law of conservation of energy
Energy at the bottom of the incline= energy at the point where the block will stop
Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=
Energy at the point where the block will stop consists of only gravitational potential energy=
Hence from Energy at the bottom of the incline= energy at the point where the block will stop
⇒
⇒
Also 
where 
is the mass of block
is acceleration due to gravity=9.8 m/s
is the difference in height between two positions
⇒
Given m=2100kg
k=22N/cm=2200N/m
x=11cm=0.11 m
∴
⇒
⇒
⇒h=0.0006467m=
They have a negative oxidation number because nonmetals gain electrons, thus making them negatively charged ions (anions). Metals become cations and have a positive charge because they lose electrons.
Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
