Answer:
Explanation:
Time dilation formula is
T = T₀ / √ 1-v²/c²
T₀ is time elapsed in moving reference , T time elapsed in stationary reference.
Here T₀ = 1 second
T = 1/√ 1-0.9² = 1/.4358 = 2.3 second
So 2.3 second will pass for each second on moving reference.
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
<h2>
A N S W E R : –</h2>
Nothing happens to the brightness of the light bulbs in the parallel circuit if the power supply is capable of supplying the additional current.
<span> The box is exerting a force of its weight. So we'll use W=mg (Note: you can use the notation F=ma, doesn't matter since weight IS a force.) We can call g negative since it's directed down towards the earth.
</span>
<span>W = 5kg(-9.8 m/(s^2)) = -49 N
</span>
<span>Thus, the box exerts a force of 49N downwards. </span>