Answer:
Approximately (assuming that .)
Explanation:
Let denote the force that this spring exerts on the object. Let denote the displacement of this spring from the equilibrium position.
By Hooke's Law, the spring constant of this spring would ensure that .
Note that the mass of the object attached to this spring is . Thus, the weight of this object would be .
Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, .
The spring in this question was stretched downward from its equilibrium by:
.
(Note that is negative since this displacement points downwards.)
Rearrange Hooke's Law to find in terms of and :
.
125 because f=ma so you would use 100=mx0.75
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For this case, the first thing you should observe is the variables:
Vertical axis: speed
Horizontal axis: time
Since the relationship between both variables is linear then we have that the slope of the line represents the acceleration according to the formula:
Where,
a: acceleration
t: time
vf: final speed
vo: initial speed
According to the equation, in the interval of 6 to 8 seconds, we observe that the slope of the Elan line is greater than the slope of the Anna line.
Therefore, for the 6 to 8 second interval, the Elan acceleration is greater.
Answer:
From 6 to 8 seconds, Elan accelerated faster than Anna.
Answer:
Explanation:
Let the required velocity of rocket be v .
We shall use the formula of time dilation to find the velocity of rocket .
t =
t = 430
t' = 38
= .0078
=.9922
= .996
v = .996 x 3 x 10⁸ m /s
= 2.988 x 10⁸ m /s
B )
Kinetic energy of rocket
= 1/2 m v²
= .5 x 20000 x (2.988 x 10⁸ )²
= 8.9 x 10²⁰ J .
C )
This energy is 8.9 times the energy requirement of United states in the year 2005 .