32 kg m/s would be the kinetic energy.
<span>Acceleration is the rate of
change of the velocity of an object that is moving. This value is a result of
all the forces that is acting on an object which is described by Newton's
second law of motion. Calculations of such is straightforward, if we are given
the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:
d = v0t + at^2/2
We cancel the term with v0 since it is initially at rest,
d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2
</span>
Explanation:
Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another
A. <span>I .................
</span>
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
