Answer:
a = - 3.75 m/s²
negative sign indicates deceleration here.
Explanation:
In order to find the constant deceleration of the car, as it stops, we will use the 3rd equation of motion. The 3rd equation of motion is as follows:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration of the car = ?
Vf = Final Velocity = 0 m/s (Since, the car finally stops)
Vi = Initial Velocity = 30 m/s
s = distance covered by the car = 120 m
Therefore,
a = [(0 m/s)² - (30 m/s)²]/(2)(120 m)
<u>a = - 3.75 m/s²</u>
<u>negative sign indicates deceleration here.</u>
<span>The kinetic energy of gas molecules is directly proportional to the Kelvin temperature of the gas</span>
Answer:
W₃ = 3310.49 J
, W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
-- If acceleration and velocity are in the same direction,
then the object is speeding up.
-- If acceleration and velocity are in opposite directions,
then the object is slowing down.
-- If acceleration is perpendicular to velocity, then the object
is moving on a circular curve at constant speed.