Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile
The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile
The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.
Answer:
The heavier piece acquired 2800 J kinetic energy
Explanation:
From the principle of conservation of linear momentum:
0 = M₁v₁ - M₂v₂
M₁v₁ = M₂v₂
let the second piece be the heavier mass, then
M₁v₁ = (2M₁)v₂
v₁ = 2v₂ and v₂ = ¹/₂ v₁
From the principle of conservation of kinetic energy:
¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J
¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400
¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400
K.E₁ + ¹/₂K.E₁ = 8400
Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁
1.5 K.E₁ = 8400
K.E₁ = 8400/1.5
K.E₁ = 5600 J
K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J
Therefore, the heavier piece acquired 2800 J kinetic energy
The scientific method. When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
Answer:
Δd = 
Explanation:
As 
, when the car is making full stop, 
. 
. Therefore,
Apply the same formula above, with 
and 
, and the car is starting from 0 speed, we have
As 
. After 
, the car would have traveled a distance of
Hence 
As 
we can simplify 
After t time, the train would have traveled a distance of 
Therefore, Δd would be 
