When you drop an object from some height on the spring objects initial potential energy is converted to kinetic energy, this kinetic energy is then used to compress the spring. Once all the energy is used the spring stops compressing and starts oscillating.
We need to find how much the spring compressed in both cases.
From the above analysis, we can conclude that potential energy in the gravitational field has to be equal to the potential energy of compressed spring.
We solve for x (in our case h=1):
Now we just have to plug in the numbers:
Answer;
velocity(v) = 97.2 m/s ,
distance (S) = 100 m
determine time(t) = ?
We know that,
distance (S) = velocity(v) × time(t)
So, time (t) = distance ÷ velocity
= 100 ÷ 97.2
<em> t = 1.02 sec.</em>
<em>I hope this will help you.</em>
The answer is C, A CD-ROM is a compact disk that contains stored
information that can be read on a computer. which can be used and store
for a long time. A CD-ROM store data in digital format. data is written
from a laser light on the disk. Minimum capacity this drive is around
700 MB.
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V = 
(1)
Total cost, 
(2)
From eqn (1):
Using the above value in eqn (1):
Differentiating the above eqn w.r.t 'a':
For the required solution equating the above eqn to zero:
a = 4
Also
The path in order to minimize the cost must be a rectangle.
