To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by
Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that
Therefore,
Re-arrange to find x,
The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.
Answer:
7557120 kg/hour
Explanation:
Given data;
Volume of air in one second = 1640 L
Density of air = 1.28 kg/L
Mass of air in 1 hour =?
Since mass = density × volume
==> Mass of air in one second = 1.28 ×1640 = 2099.2 kg
==> Mass of air in one minute = 2099.2×60=125952 kg
==> Mass of air in one hour = 125952× 60 = 7557120 kg
So rate of flow of air is 7557120 kg/hour
Answer:
m = 2.2 x 10⁻⁴ kg = 0.22 g
Explanation:
The surface tension of water is 0.072 N/m. So in order for the bug to avoid sinking, its weight per unit length of contact must be no more than the surface tension of water. Therefore,
where,
m = mass of bug = ?
g = acceleration due to gravity = 9.81 m/s²
L = Contact length = (contact length of each leg)(No. of Legs) = (5 mm)(6)
L = 30 mm = 0.03 m
Therefore,
<u>m = 2.2 x 10⁻⁴ kg = 0.22 g</u>
