We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
Answer:
a. 
b. 
Explanation:
The inertia can be find using
a.





now to find the torsion constant can use knowing the period of the balance
b.
T=0.5 s

Solve to K'


Answer:
= 829.69 Watt
≅ 830 Watt
Explanation:
Given that,
Velocity of air flow = 12.5m/s
Rate of flow of air = 9m³/s
Density of air = 1.18kg/m³
power by kinetic energy = 1/2(mv²)
mass = density × volume
m = 1.18 × 9
= 10.62 kg/s
power = 1/2 mV²
= 1/2 (10.62 × 12.5²)
= 829.69 Watt
≅ 830 Watt
Flow rate
u
=
9
m
3
/
s
Velocity of the air
V
=
8
m/s
Density of the air
ρ
=
1.18
kg
/
m
3
Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North
Explanation:
In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).
Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC
AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)
Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees
Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North
Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North
Answer:
0.214 m
Explanation:
In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.
Therefore, we can write:

where
k is the Coulomb constant
is the charge on the balloon
is the charge on the bag
r is the separation betwen the bag and the balloon
is the mass of the bag
is the acceleration due to gravity
Solving for r, we find the distance at which the bag must be held:
