Answer:
the value of x is 3.7 because they are arranged in a particular manner.
Answer: 9.98 *10^-19 J
Explanation: In order to explain this probelm we have to consider the balance enegy for photoelectric effect.
h*f-W=Ek where h is the Planck constant and W the work function and Ek the kinetic energy. f is the frequency of light.
W=h*f-Ek=6.62*10^-34*2.4*10^15-5.9*10^-19=9.98*10^-19J
Answer:
The acceleration of the plane is 10.93 m/s² at that instant.
Explanation:
Given:
Mass of aircraft, 
Downward thrust produced, 
Upward resistance force by air, 
Weight of the aircraft acting donward, 
According to Newton's second law, net force acting on an object is equal to the product of mass and its acceleration.
Here, the net force is given as the difference of downward forces and upward forces.
Now,
Therefore, the acceleration of the plane is 10.93 m/s² at that instant.
Answer:
1.86 cm
Explanation:
The center of gravity of the combined object x = m₁x₁ + m₂x₂ + m₃x₃/m₁ + m₂ + m₃ where m₁ = mass of sphere on left end of rod = 0.700 kg, x₁ = distance of center of mass of sphere from left end of combined object = radius of sphere = 8.00 cm, m₂ = mass of rod = 0.300 kg, x₂ = distance of center of mass of rod from left end of combined object = diameter of sphere on left end + length of rod/2 = 16.00 cm + 40 cm/2 = 16.00 cm + 20 cm = 36 cm, m₃ = mass of sphere on right end of rod = 0.580 kg, x₃ = distance of center of mass of sphere from left end of combined object = length of rod + diameter of first sphere + radius of sphere = 40 cm + 16 cm + 6.00 cm = 62 cm
x = (m₁x₁ + m₂x₂ + m₃x₃)/m₁ + m₂ + m₃
Substituting the values of the variables into the equation, we have
x = 0.700 kg × 8.00 cm + 0.300 kg × 36 cm + 0.580 kg × 62 cm/(0.700 kg + 0.300 kg + 0.580 kg)
x = 5.6 kg cm + 10.8 kg cm + 35.96 kg cm/1.580 kg
x = 52.36 kgcm/1.580 kg
x = 33.14 cm
Since the center of the rod is at x' = (40 cm + 16.00 cm + 12.00 cm)/2 = 70.00 cm/2 = 35 cm
The distance between the center of the rod and the center of gravity is x' - x = 35 cm - 33.14 cm = 1.86 cm
So, the center of gravity is 1.86 cm away from the center of the rod and closer to the 0.700 kg sphere.
