Each hour 430 quintillion Joules of energy from the sun hits the Earth.
In a year it is very hard to determine because of the night and different light levels.
Answer:
a = g = 9.81[m/s^2]
Explanation:
This problem can be solve using the second law of Newton.
We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.
m*g = m*a
where:
g = gravity = 9.81[m/s^2]
a = acceleration [m/s^2]
Note: If the skydiver will be under air resistance forces his acceleration will be different.
Answer:
a) 6076 m
b) 43.33 m/s
c) 68 m/s
Explanation:
(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius
s = 2r = 2* 3.38km = 6.76 km or 6760 m
(b) The average velocity would be displacement over unit of time
v = s/t = 6760 / 156 = 43.33 m/s
(c) The length of the chord it's swept in 156s is half of the circle perimeter
c = πr = π3.38 = 10.62 km or 10620 m
The airplane average speed is its chord length over a unit of time
c / t = c / 156 = 68 m/s
Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s
