Answer:
Net force that acts on the pilot = 93 N (rounded off to nearest 1 N)
Explanation:
An airplane has a mass of 3.1 × Kg
The applied force for it to take off = 3.7 × N
According to Newton's second law of motion; applied force (F) is directly proportional to the product of mass (M) and acceleration (A) of the object.
i.e F ∝ MA
This means that the applied force (F) is also directly proportional to the mass (M) of the object.
i.e F ∝ M
Acceleration (A) in this case is constant.
If the airplane's mass = 3.1 × kg requires 3.7 ×
N to take off,
Then the pilot's mass of 78 kg will require;
= 93.09677419 N = 93N (rounded off to nearest 1 N)
Answer:
The net force on the skater is zero. ()
Explanation:
According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. ()
Answer:
The value is
Explanation:
From the question we are told that
The relationship between the number of whooping cranes and the number of decades is
The exponential relationship is
Now from the given equation we have that
So comparing this equation obtained an the given exponential relationship we have that
Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar = 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar =
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar =
Total contraction on the Bar =
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm