Answer:
a) It takes her 1.43 s to reach a speed of 2.00 m/s.
b) Her deceleration is - 2.50 m/s²
Explanation:
The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:
v = v0 + a · t
Where:
v = velocty.
v0 = initial velocity.
a = acceleration.
t = time.
a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:
v = v0 + a · t
2.00 m/s = 0 m/s + 1.40 m/s² · t
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
It takes her 1.43 s to reach a speed of 2.00 m/s
b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:
v = v0 + a · t
0 = 2.00 m/s + a · 0.800 s
-2.00 m/s / 0.800 s = a
a = -2.50 m/s²
Her deceleration is - 2.50 m/s²
Well, one of the theories was that Charles darwin proved that theory evolution is real so that one theory.
Answer:
20.2 seconds
Explanation:
The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.
The crate is in free fall, so a = -9.8 m/s².
The crate falls downward, so Δx = -2000 m.
Find: t, the time it takes for the crate to land.
Δx = v₀ t + ½ at²
-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 20.2 s
It takes 20.2 seconds for the crate to land.
Answer:
S = V t where S is the horizontal distance traveled
1/2 g t^2 = H where H is the vertical distance traveled
t^2 = 2 H / g
V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations
tan theta = H / S
V^2 = S g / (2 tan theta)
Using S = L cos theta
V^2 = L g cos theta / (2 tan theta)
Giving V in terms of L and theta
<span>Cobalt-60 is undergoing a radioactivity decay.
The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span> n </span>⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.
So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567
t = 3÷0.567
= 5.290626524
the half-life of Cobalt-60 is 5.29 years.
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