Answer:
the cat is 0.4238 m in front of the dog as it leaps through the window
Explanation:
Given that;
acceleration a = 0.85 m/s²
speed v = 1.40 m/s
the cat is at rest, so initial velocity u = 0
we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;
using the second equation equation of motion;
s = ut + 1/2 at²
we substitute
2.8/2 = 0×t + 1/2 × 0.85 × t²
1.4 = 0.425t²
t = √( 1.4 / 0.425 )
t = 1.81497 sec
now, at acceleration 0.10 m/s²
the dog has to cover the distance;
s = ut + 1/2 at²
s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²
s = 2.540958 - 0.1647
s = 2.3762 m
The cant in front of the dog as it leaps through the window;
distance = 2.8 m - 2.3762 m
distance = 0.4238 m
Therefore, the cat is 0.4238 m in front of the dog as it leaps through the window
.
a phenomenon whereby the moon appears bluish owing to smoke or dust particles in the atmosphere
Your answer will be A) KILOGRAMS
Answer:
ρ = 1.13 10⁴ km/m³
Explanation:
For this exercise we use Newton's equilibrium equation
B –W + W_scale = 0
Where B is the thrust and W_scale is the balance reading
The push is given by Archimedes' law
B = ρ_water g V
B = W- W_scale
B = m g - m_scale g
Let's calculate
B = 14.7 9.8 - 13.4 9.8
B = 12.74 N
ρ_water g V = 12.74
V = 12.74 / ρ_water g
V = 12.74 / 1000 9.8
V = 0.0013 m³
Let's use density
ρ = m / V
We replace
ρ = 14.7 / 0.0013
ρ = 1.13 10⁴ km/m³
Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
