(a) The length of the train is 54.6 m
(b) The braking distance of the road train is 194.44 m
The given parameters:
acceleration of the car, a = 3.7 m/s²
initial velocity of the car, u = 30 m/s
final velocity of the car, v = 130 km/h = 36.11 m/s
To find:
The length of the train is the distance travelled by the car
The distance traveled by the car is calculated as:
Thus, the length of the train is 54.6 m
(b) The braking distance of a road train travelling at 25 m s⁻¹
Thus, the braking distance of the road train is 194.44 m
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Answer:
C
Explanation:
An object in motion will stay in motion unless acted on by a net positive or negative force.
For answer A. If the object were to be in an orbit, it would inevitably accelerate due to it being acted on by the gravitational force from the object it is orbiting. At different points in the orbit, the object will move at different speeds and continuously transfer between kinetic and potential energy.
For answer B. The object would would not stop their motion. In order for the object to lose energy, it would have to transfer it through friction or through its interaction with a gravitational field.
For answer D. No energy is "required" to maintain constant motion unless the object is willingly fighting against a resistive force like friction or a graviational well.
Answer
The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.
Explanation
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Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = 
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m 
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = 
calculate
f = 
f = 5.3 Hz
Answer:
60 rad/s
Explanation:
∑τ = Iα
Fr = Iα
For a solid disc, I = ½ mr².
Fr = ½ mr² α
α = 2F / (mr)
α = 2 (20 N) / (0.25 kg × 0.30 m)
α = 533.33 rad/s²
The arc length is 1 m, so the angle is:
s = rθ
1 m = 0.30 m θ
θ = 3.33 rad
Use constant acceleration equation to find ω.
ω² = ω₀² + 2αΔθ
ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)
ω = 59.6 rad/s
Rounding to one significant figure, the angular velocity is 60 rad/s.
