The time required by the car to stop is 4.916 sec.
Since the car is moving with the constant deceleration we can apply the first equation of motion to calculate the time required by the car to stop.
The first equation of motion is given as
V=u+at
Here, V=final speed of the car=0 mi/h as the car stops
u =initial speed of the car=55 mi/hr=24.58 m/s
a= acceleartion =-5 m/s^2 (here negative sign indicates for deceleration)
Now applying the values in the first equation
V=u+at
0=24.58-5*t
t=4.916 sec
Therefore the car will stops in 4.916 sec.
Well, half-life is the radioactivity of an identified isotope that decreases by half of the actual value.
So, your answer would be C.
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
Answer:
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- Deviance
- Dysfunction
- Distress
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